* Simplify the expression: * We can use the identities tan(x) = sin(x) / cos(x) and cosec(x) = 1 / sin(x) to rewrite the expression: sin(x) tan(x) + cos(x) cosec(x) = sin(x) * (sin(x) / cos(x)) + cos(x) * (1 / sin(x)) = sin^2(x) / cos(x) + cos(x) / sin(x) * Integrate using the formulas: * Now we can integrate each term separately using the formulas for sin(x) and cos(x): ∫ (sin^2(x) / cos(x) + cos(x) / sin(x)) dx = -∫ cos(x) dx + ∫ sin(x) dx = sin(x) - cos(x) + C Therefore, the integral of sin x tan x + cos x cosec x is sin(x) - cos(x) + C Bohot Jada Mehnat Lagi hai type karne me 😅
Change tan x into sinx/cosx and cosec x as 1/sinx then ; It becomes sin^2x / cos x + cosx /sinx then: Write sin^2x as 1-cos^2x then seprate and solve and ans will be ln|secx+tanx| + ln|sinx| - sinx + c ❤
Bruhh Pehle stanx ko sinx/cosx likho aur for sin¢2x ko 1-cos^2x likhdo jisse eqn bnjaaegi integration of secx+cosx Second part ke lie cot theta ka integration krdo Basic formulaes Ez maths
Simplifying the expression =int(sin²x/cosx + cosx/sinx)dx =int(1-cos²x/cox + cotx)dx =int(1/cosx-cos²x/cosx)dx + int(cotx)dx =int(secx-cosx) + log(sinx) + C =log(secx+tanx) -(-sinx) + log(sinx) +C And the final answer is =log{(sinx+sin²x)\cosx} + Sinx + C
@@prt-sky69bro please guide me i am new in class 11th. Which coaching should I take How should I start preparation and which books to follow? How did you prepared. Which mistakes I don't have to do
@@UdaySingh-rn4fx mathematics me ek X ki equation ke liye infinite number of values aa jati hai to ye to algebric me hai Phir bhi kaise nikala bta deta hu Sin.tan ko sin²/cos banaya aur cosec.cos ko cos/sin banaya Phir sin²/cos me sin² ko 1 - cos² se replace kiya aur cos/sin me let sin=t krke sovle kiya aur answer agya Log|sec + tan| - sin + log|sin| + c
Uss sawaal me direct product nikalo aur sin square theta ko 1-cos square theta kardo Aur dusre plus wale part me niche wale cos square theta se cos theta =t lelo phir go upar uska derivative hai phir game jaisa khelke kaam tamam kardo
To integrate the expression \( \sin x \tan x + \csc x \cos x \), we can start by rewriting it in a simpler form. 1. Rewrite \( \tan x \) and \( \csc x \): \[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \] 2. Substitute these into the expression: \[ \sin x \tan x = \sin x \cdot \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x} \] \[ \csc x \cos x = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} \] Now the integral becomes: \[ \int \left( \frac{\sin^2 x}{\cos x} + \frac{\cos x}{\sin x} ight) \, dx \] This can be separated into two integrals: \[ \int \frac{\sin^2 x}{\cos x} \, dx + \int \frac{\cos x}{\sin x} \, dx \] 3. The first integral, \( \int \frac{\sin^2 x}{\cos x} \, dx \), can be rewritten using the identity \( \sin^2 x = 1 - \cos^2 x \): \[ \int \frac{1 - \cos^2 x}{\cos x} \, dx = \int \left( \frac{1}{\cos x} - \cos x ight) \, dx = \int \sec x \, dx - \int \cos x \, dx \] 4. The second integral, \( \int \frac{\cos x}{\sin x} \, dx \), is simply: \[ \int \cot x \, dx \] 5. Now, we compute the integrals: - The integral of \( \sec x \) is \( \ln |\sec x + \tan x| + C \). - The integral of \( \cos x \) is \( \sin x + C \). - The integral of \( \cot x \) is \( \ln |\sin x| + C \). Putting it all together, we have: \[ \int \left( \sin x \tan x + \csc x \cos x ight) \, dx = \ln |\sec x + \tan x| - \sin x + \ln |\sin x| + C \] Thus, the final result is: \[ \int (\sin x \tan x + \csc x \cos x) \, dx = \ln |\sec x + \tan x| + \ln |\sin x| - \sin x + C \] 😢😢😢😢😢😢
Limit not difined hai answer alag ho sakta hai Acording to me- Int(sinxtanx)=> Int(sin²x/cosx)=> Int(1-cos²x)/cosx=> Int(1/cosx - cos²x/cosx)=> Int(secx-cosx) Int(secx) - int(cos)=> Ln|tanx +secx| - sinx+c Similarly 2nd part
§(tanx.sinx+cosecx.cosx)dx By using formula tanx=sinx/cosx Cosex=1/sinx So we can write it §(sinx. sinx /cosx+cosx .1/sinx We know that cosx/sinx=cotx and sin^2x=1-cos^2x So §(1-cos^2x/cosx+cotx)dx §secxdx-§cosxdx+§cotxdx ln|secx+tanx|+sinx+ln|sinx|+c😅
Bhai phele wala simply hojayega jab tan theta ko sinx/cosx likhoge upar 1-cos²x bach jayega fir numerator ko alag alag krlo Aajyega pehle wle ka ans Aur dusra wala part to cot theta likha hua h jiska integration baccha baccha janta h lnlsinxl Btw and is Lnlsecx+tanxl + sinx + lnlsinxl + C C is very important btw 😅😅
mera answer root log(cosx) kyu a ra hai? btw i just started integration today, (2026tard) ky ham values ko root me calc kr skte he kya? like my thing came out to be integral of root (sec x d(cosx))
Mene socha easy hai karte hai phir dekha to 1 ghanta lag gya phir jab dobara short dekha to pata chala vo second part cosec cos tha mene cotcosec le liya
My sir tells me that if you feel any topic really difficult then just solve 200 questions from that chapter. If you solve 200 question then you can solve advanced questions very easily from that particular chapter. So all the best my friends
Separate the integral sin theta*tan theta and cosec theta*cos theta And solve separately by integration by parts🤗method its so easy😅,you won't get trapped in this part ANSWER IS ln(secx+tanx)+ln(sinx)-sinx +c and i counted 1 as a constant so i ignored it,replace x with theta🤓
Bhai pehle dono terms ko alag kr lo second term simple hai ... Second term ko simplify kroge to cot x aa jayega jiska intigration log |sinx|+ C or Rahi baat first term ki to tan ki din cos me change kr lo aur sin ² upper cos niche phir sin ko 1-cos²x likho phr term aa jayega 1-cos²x / cos x . Ab dono ko alag kr lo 1/cos x = secx aur cos ²x / cos x = cos x aayega. Ab dono ko alag alag intigrate krenge cos ka sin aur sec ka log |secx + tanx|. Final answer : log|sinx|- sinx+log|secx +tanx| + C
From chat gpt To solve the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\), let's break it down into simpler parts using trigonometric identities. 1. \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) 2. \(\csc\theta = \frac{1}{\sin\theta}\) Using these identities, the expression becomes: \[ \sin\theta \left(\frac{\sin\theta}{\cos\theta} ight) + \left(\frac{1}{\sin\theta} ight) \cos\theta \] Simplify each term separately: \[ \frac{\sin^2\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \] Combine the fractions over a common denominator: \[ \frac{\sin^2\theta \sin\theta + \cos^2\theta \cos\theta}{\cos\theta \sin\theta} \] Simplify the numerator: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \] Since the numerator cannot be simplified further with common trigonometric identities, the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\) simplifies to: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \] Therefore, the simplified form of the given expression is: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \]
So answer is firsty do sepret devaid=√sinxtanxdx+√cosecxcosx and weknow sinx=cosx/cotx and tanx =sinx/cosx put in √sinxtanx then we have = √sinx/cotxdx and sinx= cosx/cotx and put in after this we have *√cosx* then we go secand part √cosecxcosxdx we know cosex =cotx/cosx put and we have *√cotx* after all thi we have √cosxdx+√cotxdx after integret this we have fainali*sinx+log[six]+c* Ungliyo me dard ho gya 😂😂😂😊
Bhaiya aap iss question ko aise solve kijiye tantheta ko sin/cos break karlo aur likh do sinsquare theta /costheta then sintheta ko t man lo phir aapka expression t^2dt aa jayega phir udhar aap cosec theta ko 1/sintheta kardo wo banjayega cot theta aur ha pehle integration ko break karlena aapka answer phir ayega sin^3theta/3+ln sintheta +c 😊
Sir woh 2nd part easily ho jata hai.... Sirf first part ko sin and cos ke terms me likho aur fir take cos as t fir integration by parts... fir (1-t²) ko v manlo
First seprate the integral the second term turs out to be cotθ which is a standard integral the first term turns out to be (sinθ)^2/cosθ use identity (sinθ)^2= 1-(cosθ)^2 then divide and get two standard integrals
dono ko alag alag integrate krlo right side wala to straight up {cotx} aur left side wala {secx} aur {-cosx} ban jayega ab bas jo bracket me hai unhe integrate krlo ans ln|secx+tanx| -sinx+log|sinx| aajaega.
Use by part Best method to solve integration is remember 15. Ways or type of inte. Method which ur good teacher gave u in notes Aur xam m 1 min phir last m inte ka ques ko time deena
I can do any hard question but if my friends give me cbse questions i cant do due to nervousness and self respect and while alone (no flexing) can solve very hard questions
The correct answer to this question is ln(tanx+secx)+ln(sinx)-sinx+C we are taking theta is equal to x here since I don't know how to write theta in keyboard
*∫(sin(x)tan(x) + cosec(x)cos(x)) dx *We can simplify the integral by using trigonometric identities: *1. tan(x) = sin(x)/cos(x), so sin(x)tan(x) = sin^2(x)/cos(x) *2. cosec(x) = 1/sin(x), so cosec(x)cos(x) = cos(x)/sin(x) ∫(sin^2(x)/cos(x) + cos(x)/sin(x)) dx u = cos(x), du/dx = -sin(x) *The integral becomes: ∫(-sin(x)(sin^2(x)/cos(x) + cos(x)/sin(x))) dx = ∫(-sin^3(x)/cos(x) - cos^2(x)/sin(x)) dx = ∫(-u^2/u - (1-u^2)/u) du (substituting u = cos(x)) Now, we can integrate: = ∫(-u - 1/u + u) du = ∫(-1/u) du = -ln|u| + C = -ln|cos(x)| + C So, the final answer is: ∫(sin(x)tan(x) + cosec(x)cos(x)) dx = -ln|cos(x)| + C *Where C is the constant of integration. So guys this is the correct answer with correct method🥀
Question incomplete hai integral ka sign bas sin theta + cos theta ke aage laga hai😂😂 Kiske respect mai karna hai wo bhi nahi pata 😂 Bonus milna chahiye bhai ko💀💀💀💀
Solution: 1.Write the expression in terms of sinx and cosx. 2.Convert all sinx into cosx using [(sinx)^2 +(cosx)^2 =1] 3.Divide and break the fraction (a-b/c =a/c-b/c) Remaining: secx-cosx+cosecx.cotx Now integrate... Answer: ln(|secx+tanx|) - sinx - cosecx + c
First rule of integration: kabhi bhi koi bhi question ese hi shuru na kardo pehle uski validation Karo , tabhi shuru Karo , ho sakta hai wo non integrable ho... *Agar sawaal na aaye to ye gyaan pel ke patli gali se nikal lo.😂
For sinx tanx = break in sin cos, 1-cos sqaure/ sin , take sin t and alag alg solve khatam Now the next funtion break in sin cos, sin se multiply upr niche, 2sixcos=dt and sin sqaure x=t Khatam
![เป็นไปได้ไหม - WanMai [Performance Video]](http://i.ytimg.com/vi/Eh5hItD4Mr0/mqdefault.jpg)
Aadhe logo ko joke hi smjh ni aaya 😂
Secx.tanx - sinx + log|sinx| + c.
Aadhe logo ko joke hi ni smjh aaya 😂
Are ye sab toh toh chhote mote target hai aap to 10M deserve karte ho
@@asxff730galat hai log|tanx + tanx sinx| - sinx hoga
ln|secx+tanx|-sinx-ln|sinx|+c
* Simplify the expression:
* We can use the identities tan(x) = sin(x) / cos(x) and cosec(x) = 1 / sin(x) to rewrite the expression:
sin(x) tan(x) + cos(x) cosec(x) = sin(x) * (sin(x) / cos(x)) + cos(x) * (1 / sin(x))
= sin^2(x) / cos(x) + cos(x) / sin(x)
* Integrate using the formulas:
* Now we can integrate each term separately using the formulas for sin(x) and cos(x):
∫ (sin^2(x) / cos(x) + cos(x) / sin(x)) dx = -∫ cos(x) dx + ∫ sin(x) dx
= sin(x) - cos(x) + C
Therefore, the integral of sin x tan x + cos x cosec x is sin(x) - cos(x) + C
Bohot Jada Mehnat Lagi hai type karne me 😅
Respect 👍🏻
Fir bhi glt javab aaya
😂
Chat GPT Zindabaad
Ans fir bhi galat hai bro, tune kaise integrate kar diya
Mera to Log(sec(θ) + tan(θ)) - sin(θ) + log(sin(θ)) aaya
Ye 2 saal wale bhaiya me mujhe apna chehra kyun dikh raha hai😅😅😅😅😅
Us bro us 😢😢
Same bhai 😅😅
kyuki ye reel bana raha hai aur to dekh raha hai😂😂😂😂😂😂
bhai tumhara naam..............
🙂
No matter where u study in the future, integration is always terrifying 😂
Put that question equal to 'I' now solve 'I'
😂😂😂😂😂😂
Sin²x/cosx + cotx
So sin²x = 1-cos²x
Therefore a new expression is (secx+cosx+cotx)dx
Ab to ye baccho ka khel hai koi bhi solve kar lega
Arein pata hain bhai ....kyu Gyan jhaad raha hain@@rudranshdixit9184
@@rudranshdixit9184
Kon sa baccha integration khelta hai😢
Bhai jab itna bataya to thoda aur bata deta@@rudranshdixit9184
@@rudranshdixit9184bro mera toh integral(sec∅ + cosec∅cot∅ + cos∅)d∅ aarha hai jiska solution
ln|sec∅ + tan∅| -cosec∅ -sin∅ + C aarha h
Preparation ❎
Ratta ☑️
Change tan x into sinx/cosx and cosec x as 1/sinx then ;
It becomes sin^2x / cos x + cosx /sinx then:
Write sin^2x as 1-cos^2x then seprate and solve and ans will be
ln|secx+tanx| + ln|sinx| - sinx + c ❤
Yes itne easy Q me sab bakchodi kar rahe hai
@@upprrdimensions892 ha bhai
when you realize, this is a halwa integration qn
Yes integration by parts lagega🤓🤓
No substitution lagega
@@NITianpramod bhai nahi to tum pahle hi simplify kardo integration ko easily solve ho jaayega ,substitution aur ibp mein time lagega
Yes fr
@Vayu_Aksh no bro
Ans - ln|sinx|+ln|secx+tanx|-sinx+C
Kitne saal hogye taiyari karte hue
Wrong
Sin x + Log l sinx l + C
Isko aise bhi kr sakte ln|1-cos@|-sin@
If it had to be integrated wrt to d@
@@lostbeatz8782sahi hai mera bhi wahi aaya hai..!
Sahi ans hai
ln| secx+tanx| + ln|sinx| -sinx + C
Aur last me x ke jagah theata rakh Dena 👍
kar mila ye karke job mil gai?
@@abhinavgupta2885satisfaction ki tuzse ye nhi ho rha ye dekhkar
❤
Tumhe ye comment dekh kar jal raha hai?🤣@@abhinavgupta2885
@@abhinavgupta2885sukoon
Bruhh
Pehle stanx ko sinx/cosx likho aur for sin¢2x ko 1-cos^2x likhdo jisse eqn bnjaaegi integration of secx+cosx
Second part ke lie cot theta ka integration krdo
Basic formulaes
Ez maths
Simplifying the expression
=int(sin²x/cosx + cosx/sinx)dx
=int(1-cos²x/cox + cotx)dx
=int(1/cosx-cos²x/cosx)dx + int(cotx)dx
=int(secx-cosx) + log(sinx) + C
=log(secx+tanx) -(-sinx) + log(sinx) +C
And the final answer is
=log{(sinx+sin²x)\cosx} + Sinx + C
Ans> Log(secx + tanx) - sinx + log(sinx) +c
Bhaiya ans khali log(secx+tanx) +c hoga bs
@@atharvmishra2810 nahi bhai apna answer recheck karo.
BTW konsa saal hai prep ka?
@@prt-sky69bro please guide me i am new in class 11th. Which coaching should I take
How should I start preparation and which books to follow? How did you prepared. Which mistakes I don't have to do
Comments me Dusre wale bhaiya toh kuch aur answer bata rahe hai😂😂😂
@@UdaySingh-rn4fx mathematics me ek X ki equation ke liye infinite number of values aa jati hai to ye to algebric me hai
Phir bhi kaise nikala bta deta hu
Sin.tan ko sin²/cos banaya aur cosec.cos ko cos/sin banaya
Phir sin²/cos me sin² ko 1 - cos² se replace kiya aur cos/sin me let sin=t krke sovle kiya aur answer agya
Log|sec + tan| - sin + log|sin| + c
The answer to this question is ln(secx + tanx / sinx) - sinx + c
Bhai ans galat hai phir se chake kar lo
Kyoki sin x भाग मे नही गुडा मे रहेगा क्योकि प्लस मे log sin x hai
Bro x ke jagah theta to tune bdla hi mhi
Bhai vo bhi glat hai ans mai log|secx+tanx|-sinx-1/sinx +c hoga
Integration by parts??
Sahi ans hai
ln| secx+tanx| + ln|sinx| -sinx + C
Aur last me x ke jagah theata rakh Dena 👍
Uss sawaal me direct product nikalo aur sin square theta ko 1-cos square theta kardo
Aur dusre plus wale part me niche wale cos square theta se cos theta =t lelo phir go upar uska derivative hai phir game jaisa khelke kaam tamam kardo
Calculus and vector algebra was my fav chapter ❤
☠️☠️
Nice taste in maths.
Nice
💀
💀💀💀
Answer is:- ln | sec theta ( 1 + cosec theta ) | - sin theta + C
C IS CONSTANT OF INTEGRATION OR YOU CAN SAY
YOU ARE THAT “C”
Bro d theta is not given
Answer Is Mitochondria ☠️.
To integrate the expression \( \sin x \tan x + \csc x \cos x \), we can start by rewriting it in a simpler form.
1. Rewrite \( \tan x \) and \( \csc x \):
\[
\tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x}
\]
2. Substitute these into the expression:
\[
\sin x \tan x = \sin x \cdot \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x}
\]
\[
\csc x \cos x = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x}
\]
Now the integral becomes:
\[
\int \left( \frac{\sin^2 x}{\cos x} + \frac{\cos x}{\sin x}
ight) \, dx
\]
This can be separated into two integrals:
\[
\int \frac{\sin^2 x}{\cos x} \, dx + \int \frac{\cos x}{\sin x} \, dx
\]
3. The first integral, \( \int \frac{\sin^2 x}{\cos x} \, dx \), can be rewritten using the identity \( \sin^2 x = 1 - \cos^2 x \):
\[
\int \frac{1 - \cos^2 x}{\cos x} \, dx = \int \left( \frac{1}{\cos x} - \cos x
ight) \, dx = \int \sec x \, dx - \int \cos x \, dx
\]
4. The second integral, \( \int \frac{\cos x}{\sin x} \, dx \), is simply:
\[
\int \cot x \, dx
\]
5. Now, we compute the integrals:
- The integral of \( \sec x \) is \( \ln |\sec x + \tan x| + C \).
- The integral of \( \cos x \) is \( \sin x + C \).
- The integral of \( \cot x \) is \( \ln |\sin x| + C \).
Putting it all together, we have:
\[
\int \left( \sin x \tan x + \csc x \cos x
ight) \, dx = \ln |\sec x + \tan x| - \sin x + \ln |\sin x| + C
\]
Thus, the final result is:
\[
\int (\sin x \tan x + \csc x \cos x) \, dx = \ln |\sec x + \tan x| + \ln |\sin x| - \sin x + C
\]
😢😢😢😢😢😢
Others : padai ka video
Me: ye bhi badminton khelta wo bhi yonex ke mehenge wale racket se😂😂
Limit not difined hai answer alag ho sakta hai
Acording to me-
Int(sinxtanx)=>
Int(sin²x/cosx)=>
Int(1-cos²x)/cosx=>
Int(1/cosx - cos²x/cosx)=>
Int(secx-cosx)
Int(secx) - int(cos)=>
Ln|tanx +secx| - sinx+c
Similarly 2nd part
aur by prts ki jagah hm direct sinx tanx ko ILATE se bhi kr sakte h but usse answer differ aayega
§(tanx.sinx+cosecx.cosx)dx
By using formula tanx=sinx/cosx
Cosex=1/sinx
So we can write it
§(sinx. sinx /cosx+cosx .1/sinx
We know that cosx/sinx=cotx
and sin^2x=1-cos^2x
So
§(1-cos^2x/cosx+cotx)dx
§secxdx-§cosxdx+§cotxdx
ln|secx+tanx|+sinx+ln|sinx|+c😅
import java.util.Scanner;
public class TrigonometryCalculator {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// Input angle in degrees
System.out.print("Enter angle in degrees: ");
double angleInDegrees = sc.nextDouble();
// Convert the angle to radians as Java trigonometric functions use radians
double angleInRadians = Math.toRadians(angleInDegrees);
// Calculate sine, cosine, and tangent values
double sinValue = Math.sin(angleInRadians);
double cosValue = Math.cos(angleInRadians);
double tanValue = Math.tan(angleInRadians);
// Output the trigonometric values
System.out.println("Sine of " + angleInDegrees + " degrees = " + sinValue);
System.out.println("Cosine of " + angleInDegrees + " degrees = " + cosValue);
System.out.println("Tangent of " + angleInDegrees + " degrees = " + tanValue);
// Handle special cases where tangent is undefined (90, 270, etc.)
if (cosValue == 0) {
System.out.println("Tangent of " + angleInDegrees + " degrees is undefined (division by zero).");
}
sc.close();
}
}
To integrate the expression sin(x) × tan(x) + cosec(x) × cos(x), we can use the following steps:
1. Recall the trigonometric identities:
- tan(x) = sin(x) / cos(x)
- cosec(x) = 1 / sin(x)
2. Rewrite the expression using these identities:
- sin(x) × tan(x) = sin(x) × (sin(x) / cos(x)) = sin²(x) / cos(x)
- cosec(x) × cos(x) = (1 / sin(x)) × cos(x) = cos(x) / sin(x)
3. Combine the two expressions:
- sin²(x) / cos(x) + cos(x) / sin(x)
4. Simplify the expression:
- (sin³(x) + cos²(x)) / (cos(x) × sin(x))
5. Recall the Pythagorean identity:
- sin²(x) + cos²(x) = 1
6. Substitute this into the expression:
- (1 - cos²(x) + cos²(x)) / (cos(x) × sin(x))
7. Simplify:
- 1 / (cos(x) × sin(x))
8. Integrate:
- ∫(1 / (cos(x) × sin(x))) dx
This integral can be solved using substitution or integration by parts.
My ans is ln|secx+tanx|-sinx+ln|sinx|+c using a simple trigonometric formula and standard integration formulas
Correct ✅ ☺️
Yup easy one
Integration solve karte hain:
\[
\int (\sin x \cdot \tan x + \cos x \cdot \csc x) \, dx
\]
Sabse pehle, \(\tan x\) aur \(\csc x\) ko unke trigonometric equivalents me convert karte hain:
\[
\int (\sin x \cdot \frac{\sin x}{\cos x} + \cos x \cdot \frac{1}{\sin x}) \, dx
\]
Simplify karte hain:
\[
\int \left( \frac{\sin^2 x}{\cos x} + \frac{\cos x}{\sin x}
ight) \, dx
\]
Ab isse do alag-alag integrals me split karte hain:
\[
\int \frac{\sin^2 x}{\cos x} \, dx + \int \frac{\cos x}{\sin x} \, dx
\]
\[
\int \frac{\sin^2 x}{\cos x} \, dx + \int \cot x \, dx
\]
Pehle integral ko simplify karte hain:
\[
\int \frac{\sin^2 x}{\cos x} \, dx = \int \left( \frac{1 - \cos^2 x}{\cos x}
ight) \, dx = \int \left( \frac{1}{\cos x} - \cos x
ight) \, dx
\]
\[
= \int \sec x \, dx - \int \cos x \, dx
\]
Ab integrals solve karte hain:
\[
\int \sec x \, dx - \int \cos x \, dx + \int \cot x \, dx
\]
\[
= \ln|\sec x + \tan x| - \sin x + \ln|\sin x| + C
\]
Yeh final answer hai:
\[
\int (\sin x \cdot \tan x + \cos x \cdot \csc x) \, dx = \ln|\sec x + \tan x| - \sin x + \ln|\sin x| + C
\]
Jahan \(C\) integration constant hai.
Jise samjh me aa gya ho like kr de 😅
Use integration by parts ;
2nd function:-tan=sin/cos
1st function :-sin
UV method is also applicable
Right
Bhai phele wala simply hojayega jab tan theta ko sinx/cosx likhoge upar 1-cos²x bach jayega fir numerator ko alag alag krlo
Aajyega pehle wle ka ans
Aur dusra wala part to cot theta likha hua h jiska integration baccha baccha janta h
lnlsinxl
Btw and is
Lnlsecx+tanxl + sinx + lnlsinxl + C
C is very important btw 😅😅
This integration: 😂😂
But
Integration of √tanx : ☠️☠️
Solve this
mera answer root log(cosx) kyu a ra hai?
btw
i just started integration today,
(2026tard)
ky ham values ko root me calc kr skte he kya?
like my thing came out to be integral of root (sec x d(cosx))
@@SuccessfulInsaanTObe no wrong ans, again try
@@subhajitnath9654 im very new to integration, can u help me out?
1/2√tanx differentiation
Kal iska answer puchkar aaunga
ln|secx+tanx|-sinx+ln|sinx| +C
Maturity is when u realise ques was halwaaa😂😂
Mene socha easy hai karte hai phir dekha to 1 ghanta lag gya phir jab dobara short dekha to pata chala vo second part cosec cos tha mene cotcosec le liya
My sir tells me that if you feel any topic really difficult then just solve 200 questions from that chapter. If you solve 200 question then you can solve advanced questions very easily from that particular chapter. So all the best my friends
ans :
ln|secx + tanx|+ln|cosx|+ln|cosec2x-cot2x| - sinc +c
Wrong
Separate the integral sin theta*tan theta and cosec theta*cos theta
And solve separately by integration by parts🤗method its so easy😅,you won't get trapped in this part
ANSWER IS ln(secx+tanx)+ln(sinx)-sinx +c and i counted 1 as a constant so i ignored it,replace x with theta🤓
Bhai pehle dono terms ko alag kr lo second term simple hai ... Second term ko simplify kroge to cot x aa jayega jiska intigration log |sinx|+ C or Rahi baat first term ki to tan ki din cos me change kr lo aur sin ² upper cos niche phir sin ko 1-cos²x likho phr term aa jayega 1-cos²x / cos x . Ab dono ko alag kr lo 1/cos x = secx aur cos ²x / cos x = cos x aayega. Ab dono ko alag alag intigrate krenge cos ka sin aur sec ka log |secx + tanx|. Final answer : log|sinx|- sinx+log|secx +tanx| + C
galat its -sinx not +sinx
From chat gpt
To solve the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\), let's break it down into simpler parts using trigonometric identities.
1. \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
2. \(\csc\theta = \frac{1}{\sin\theta}\)
Using these identities, the expression becomes:
\[
\sin\theta \left(\frac{\sin\theta}{\cos\theta}
ight) + \left(\frac{1}{\sin\theta}
ight) \cos\theta
\]
Simplify each term separately:
\[
\frac{\sin^2\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}
\]
Combine the fractions over a common denominator:
\[
\frac{\sin^2\theta \sin\theta + \cos^2\theta \cos\theta}{\cos\theta \sin\theta}
\]
Simplify the numerator:
\[
\frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta}
\]
Since the numerator cannot be simplified further with common trigonometric identities, the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\) simplifies to:
\[
\frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta}
\]
Therefore, the simplified form of the given expression is:
\[
\frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta}
\]
This question answer is log|secQ +tanQ| -sinQ +log|sinQ|+c
Ans: ln|secx+tanx|-sinx-ln|cosecx|+C
This is the simplest question of integration...
ln|secx+tanx|+ln|sinx|-sinx+C
Absolutely correct
@@harshitvarshney4997yeh 12th mei ata h ya 11 th kyuki meri 11th khtm hogyi aur mujhe nhi ata plzz batado bhai😢
@@JapjotSingh-fi2it12th me 7 chapter integration me h, integration by parts, don't worry, it's easy to learn
@@RahulSharma-gv3ng thankss😅🤧
@@RahulSharma-gv3ng aur aap kya kr rhe ho 12 th mei ho?
So answer is firsty do sepret devaid=√sinxtanxdx+√cosecxcosx and weknow sinx=cosx/cotx and tanx =sinx/cosx put in √sinxtanx then we have = √sinx/cotxdx and sinx= cosx/cotx and put in after this we have *√cosx* then we go secand part √cosecxcosxdx we know cosex =cotx/cosx put and we have *√cotx* after all thi we have √cosxdx+√cotxdx after integret this we have fainali*sinx+log[six]+c*
Ungliyo me dard ho gya 😂😂😂😊
Iska answer hai log|secx+tanx| +log|Sinx| - Sinx
+ C
Or
log|secx +tanx|×sinx - sinx ( because log a + log b = log a×b)
Achha lage to please like ❤❤😊😊
Joh integration padh raha hai usko bhi toh atleast 1.5 yrs ho gaye
Bhaiya aap iss question ko aise solve kijiye tantheta ko sin/cos break karlo aur likh do sinsquare theta /costheta then sintheta ko t man lo phir aapka expression t^2dt aa jayega phir udhar aap cosec theta ko 1/sintheta kardo wo banjayega cot theta aur ha pehle integration ko break karlena aapka answer phir ayega sin^3theta/3+ln sintheta +c 😊
Bhi tumhari bat suna ki background music ki bat
Lamba ha madam lamba😂😂😅😅
Quite lengthy but easy question 😂
Solve in 5 min for first time
Solution: ❎
Song:✅
Answer is
∫(sinθ tanθ + cosecθ cosθ) dθ = -cosθ + sinθ + C
Where C is the constant of integration.
Sorry to say if you are preparing for 2 years and still couldn’t solve this sorry but you will face failure😢….
True
Simplest question🙋😊 I like solving such integration questions😊
Abe ooi meri personal life kyu share kar rahe ho yrr mujhse puche bina case kar dunga 😂😂😂😂😂😂😂😂😂😂😂
Sir woh 2nd part easily ho jata hai....
Sirf first part ko sin and cos ke terms me likho aur fir take cos as t fir integration by parts... fir (1-t²) ko v manlo
Easiest question 😂😂.
Jee mai aisa aaye toh mja aa jaye dekhte hi ho jaarha pr sadly jee main itna tough aata
Yeahh I learnt integration today only and I could solve this. I'm in class 11
@@alikarol4900 yeah me too i am in class 9
@@Villain_Arc_Has_Started chal be
First seprate the integral the second term turs out to be cotθ which is a standard integral the first term turns out to be (sinθ)^2/cosθ use identity (sinθ)^2= 1-(cosθ)^2 then divide and get two standard integrals
Itna bi mushkil sawal nhi h ye 😂😂😂
dono ko alag alag integrate krlo right side wala to straight up {cotx} aur left side wala {secx} aur {-cosx} ban jayega ab bas jo bracket me hai unhe integrate krlo ans ln|secx+tanx| -sinx+log|sinx| aajaega.
Bro iska integration nahi hoga bcoz (dx) nhi he
Bro dtheta
Un dono ko alag alag seperate kardo change into sin and cos form, pehle wale mein by parts laga do simply and second will change into ln form :)
Wo sab htao bhai ques me d theeta to hai hi nhi 😂😂 integerate kiske respect me kre
😂😂😂😂
Common sense h bhai d theta sei krega aur kisse krega
@@dakshmehta4498 hm wo hai but mathematically to ques glt hi hua na 😂
D theta valuable hai broo
@@dakshmehta4498exam me common sence nhi chalta marks katte he
Use by part
Best method to solve integration is remember 15. Ways or type of inte. Method which ur good teacher gave u in notes
Aur xam m 1 min phir last m inte ka ques ko time deena
Orally kar liya 😂😂😂
ln(secx+tanx) -sinx +ln(sinx) + c
Rank kitni aayi bhai
@@bharatbihani6184 91k mains m 😓
I can do any hard question but if my friends give me cbse questions i cant do due to nervousness and self respect and while alone (no flexing) can solve very hard questions
Question was easy😂😂
Easy to hai
1st term ko (1- cos sq)/cos
2nd term ko cot
Then sec - cos + cot ban jayega jo ki standard hai
Ans : Ln|sec + tan| - sin + ln|sin| + ❤
Bhai integration se hard toh A.P G.P aur Binomial ke questn aate hai😂💀
Bhai kahi tum PNC aur probability ko bhul to nhi gaya h😅
@@ram.......111 sab hi hard
The correct answer to this question is ln(tanx+secx)+ln(sinx)-sinx+C we are taking theta is equal to x here since I don't know how to write theta in keyboard
Neet student hoke ye question solve karne ka ghamand hai😎😎
Same here 😅
Aur hame Jee student hoke na karne ka ghammand h😂😂😂😂
Iska ghamand hai
Lekin chaar saal drop lene ka ghamand nhi hai
Maha easy question hai
10th mai solve karne ka ghamand hai😂😂
Motion Learning Aap Hai Na Sbhi Problem ha Solution '♥✨
Apply UV rule
UV = V × du/dx + U×dv/dx
Apply this on both equation and then integrate it you will get ur desired answer
Bhai ye differentiation main work karta he yahan uv ka dusra rule hota he
Abe ye differentiation kyun krr rha bhai 😂😂😂
My mistake maine notice nhi kiya
Lekin tum isi process se individually integrate krdo na aa jaayega
By using trigonometry identity
Final ans is log(tanx+secx+sinx)-sinx+c
bro be sniffin em maths problems 💀
Bro you chose the wrong comment section people here won't understand the humour 😭
@@neerubalamakeover9097 i didn't commented for the sake of attention, i genuinely felt too do so, and your point is true as well
Ans=log|secx+tanx|/log|cosecx+cotx| +cosx-sinx
Aur last me x ke jagha tita raklo ho gaya answer.....
Bhai dx kaha hee😂😂
dtheta
*∫(sin(x)tan(x) + cosec(x)cos(x)) dx
*We can simplify the integral by using trigonometric identities:
*1. tan(x) = sin(x)/cos(x), so sin(x)tan(x) = sin^2(x)/cos(x)
*2. cosec(x) = 1/sin(x), so cosec(x)cos(x) = cos(x)/sin(x)
∫(sin^2(x)/cos(x) + cos(x)/sin(x)) dx
u = cos(x), du/dx = -sin(x)
*The integral becomes:
∫(-sin(x)(sin^2(x)/cos(x) + cos(x)/sin(x))) dx
= ∫(-sin^3(x)/cos(x) - cos^2(x)/sin(x)) dx
= ∫(-u^2/u - (1-u^2)/u) du (substituting u = cos(x))
Now, we can integrate:
= ∫(-u - 1/u + u) du
= ∫(-1/u) du
= -ln|u| + C
= -ln|cos(x)| + C
So, the final answer is:
∫(sin(x)tan(x) + cosec(x)cos(x)) dx = -ln|cos(x)| + C
*Where C is the constant of integration.
So guys this is the correct answer with correct method🥀
answer ln|secx+tan x| - Sin x + ln |Sin x | + C
Yess this is correct I got the same answer 🥰
Yes i did it orally 😂😂😂
I did it oralyy 🥰🥰🥰🥰
@minakumari7552 nhi sahi hai ... aap dekho aap kahi galti kar rhe ho
Desclaimer:
Only for 12th or above math students 😊
ANSWER:- ln
(| tan(x) + sec(x) |) + ln(| sin(x) |)−sin(x)+C
Everyones calculating and get ans me how they get because bar bar multiplication ke karan integration vapis repeat kar rahai or ruk hi nahi raha😅😂
Question incomplete hai integral ka sign bas sin theta + cos theta ke aage laga hai😂😂
Kiske respect mai karna hai wo bhi nahi pata 😂
Bonus milna chahiye bhai ko💀💀💀💀
dtheta Maine lga Diya bhai khush
@@hiphopfreek6907 phir bhi incomplete hai bhai
Just use integration by parts or you can also subsitute x=pi -theta and solve fromthere
Solution:
1.Write the expression in terms of sinx and cosx.
2.Convert all sinx into cosx using [(sinx)^2 +(cosx)^2 =1]
3.Divide and break the fraction (a-b/c =a/c-b/c)
Remaining: secx-cosx+cosecx.cotx
Now integrate...
Answer: ln(|secx+tanx|) - sinx - cosecx + c
Upon simplification, It is integral of secx . We can write is as log|secx + tanx| - ln|sinx| - sinx + c
Wrong
saaheb ji ham janat hai ki ham galat hai, bas batana bhul gye
@@druhindatta1976 it's not integral of secx though, that's what I was trying to say 👍🏻🔥
@@randomanonymous-x7i got it brother
First rule of integration: kabhi bhi koi bhi question ese hi shuru na kardo pehle uski validation Karo , tabhi shuru Karo , ho sakta hai wo non integrable ho...
*Agar sawaal na aaye to ye gyaan pel ke patli gali se nikal lo.😂
It's too easy, the answer is
ln|secθ+tanxθ|+ln|sinθ|-sinθ+C, C= integral constant
Bhai kyu IIT Naam kharab kar rahe ho 😅😅😂😂
For sinx tanx = break in sin cos, 1-cos sqaure/ sin , take sin t and alag alg solve khatam
Now the next funtion break in sin cos, sin se multiply upr niche, 2sixcos=dt and sin sqaure x=t
Khatam
Bhai second function ke liye sirf sinx=t lelenge fir upar toh cos Hae hi
Bhai sabse easy hai yeah toh 😂😂
Bro me maths ka board dekar aa chuka hu tab bhi yeh mere se nhi ho rha ☠️🥲 btw me to commerce student hu mujhe kya matlab
agr question dimaag main bahut dourh rha hai toh uss x ko bara vala X manke uspr chinta baitha do...question rest main aa jayega..phir solve kr lena
Question❓
∫ { sin(x)•tan(x) + cos(x)•cosec(x) }dx
Solutions : -
Let ɪ = ∫ { sin(x)•sin(x)/cos(x) + cos(x)•1/sin(x) }dx
= ∫ { sin²(x)/cos(x) + cos(x)/sin(x) }dx
= ∫ [ {1-cos²(x)}/cos(x) + cot(x) ]dx
= ∫ [ {1/cos(x) - cos²(x)/cos(x)} + cot(x) ]dx
= ∫ [ {sec(x) - cos(x)} + cot(x) ]dx
Now Integrate : -
= log |sec(x)+tan(x)| - sinx + log |sin(x)| + C
= log | {sec(x)+tan(x)}•sin(x) | - sin(x) + C
= log | {1/cos(x)+sin(x)/cos(x)}•sin(x) | - sin(x) + C
= log | [{1+sin(x)}/cos(x)]•sin(x) | - sin(x) + C
= log | {1+sin(x)}•sin(x)/cos(x) | - sin(x) + C
∴ ɪ = log | {1+sin(x)}•tan(x) | - sin(x) + C
This is the Final Answer. 😊
Full chatgpt copy paste😂
It was very easy . Just split the integration into two small integrals and integrate individually
ln|secX+tanx|+ln|sinx|+cos2x/4 ,,, is the wright answer
First write right right
Right answer is ln(tan(x) + sin(x)tan(x)) -sin(x) + c
Ln|secx+tanx|-sinx+ln|sinx|
Ans: sinx tanx + cosecx cosx
Becaue not given wrt which integration is to be done
log(tan ◇ + sin ◇ tan ◇) - sin ◇ +c is the required answer.
When you realise this is "chindi' question of trignometry😅😅
Assan sawal hai bhaiya
Separate krke solve kro phir tan ko sin ke form cosec ko bhi usi form mein likhe substitute krdo
Ans is log|tanx(1+sinx)|-sinx+c,[here x is theta]
ln∣sec(x)+tan(x)∣ − sin(x) + ln∣sin(x)∣ + c ye ayega answer, mera to ye aaya
Last me x ko theta se replace kar dena, x likhne ki aadat ho gayi hai😅😅😅
Just write tan x as sin x/cos x then put sin^2x = 1-cos^2x then seperate as secx and cos x.
∫(sinx*tanx + cosecx .cosx)dx
Apply by parts in first intgral
tanx(-cosx)-∫ sec²x. (-cosx)dx + ∫ cotx dx
note: cosecx. cosx= cotx
-sinx + ∫ secxdx + lnsinx
-sinx + ln(secx + tanx) + lnsinx + C