Thank you for your support! We cannot apply integration by part because both functions are trigonometric functions. FYI, integration by parts is used for product of 2 different categories of functions, like polynomial with trigo and etc
for ques- 9b, why did you use substitution. In our syllabus, it was stated that Substitution will always be given when we need to use it. Here, in the question, it wasn't given.
Substitution is one of methods to integrate sinxcos2x and also the easiest method. Although u is not given in this question, but we are aware the substitution is one of the ways to solve the integration problem. d/dx ( cosx) = -sinx, hence we may write all in terms of u and get a simpler form before perform the integration
The final answer shown in the video is equivalent to the form you've mentioned. To get your form, just multiply my final answer with e^((1/x)-1) to the numerator and denominator
Hi, there are 2 methods to solve that question, i.e. graph and use definition of modulus function, then followed by squaring both sides to eliminate the square root. The checking part is to make sure the solutions satistify the inequality as there is only one modulus function in the inequality.
Modulus of 14 is 14. For your info, modulus value is always positive, so the modulus of negative value is positive. For the modulus of positive value, it is definitely a positive.
We always use OR and not AND, because as long as it satisfies 1 of the results,, then the inequality is valid. If you write AND, that means the inequality is only valid when satisfy both results
Use substitution method to solve the multiplication of same type of functions. In this case, sin and cos are both trigo and we cannot simplify them, hence use substitution method. U is the function with power but excluding the power
No, integration by parts is to integrate the product of 2 different types of functions. This is rational function involves polynomial in both numerator and denominator, hence, either ln, partial fractions or substitution method
if only i found this channel earlier, not 16 hours before my maths p3 😭
us
Good luck in your exam!
Well be good gang 🙏🏽
Us 😔😔
p3 is easy man
thank you for the effort u put in. Also, in q8, can we use integration by parts instead of substitution?????
Thank you for your support! We cannot apply integration by part because both functions are trigonometric functions. FYI, integration by parts is used for product of 2 different categories of functions, like polynomial with trigo and etc
@@MathWorld-yp9odI see, thank you for the explanation!
Thank you so much I had trouble in the last question and now it is solved.
You're most welcome. Please continue supporting this channel and stay tuned for more upcoming videos. Thank you
Thank you ❤❤❤❤❤❤❤❤❤
😉
Watching this 57 min before my P3 lol
Good luck
The amount of adds in your videos 😭😭🙏🏻🙏🏻🙏🏻
i dint even hv a single one💀
those aren't ads
for ques- 9b, why did you use substitution. In our syllabus, it was stated that Substitution will always be given when we need to use it. Here, in the question, it wasn't given.
Substitution is one of methods to integrate sinxcos2x and also the easiest method. Although u is not given in this question, but we are aware the substitution is one of the ways to solve the integration problem. d/dx ( cosx) = -sinx, hence we may write all in terms of u and get a simpler form before perform the integration
Thank you so much❤ you are literally saving my exams and making everything look easy😂
Thanks for your compliment! Good luck!
thank you !
You're most welcome. Thank you for your support and please share this channel with your friends. Stay tuned for more upcoming videos 😊
in Q10 part a why is square root of c^2 not c? the square root will cancel the power right?
We cannot cancel the power because it is square root of 20+c^2. If only square root of c^2, then we can simplify it to c
In the marking scheme for number 11(a) the answer is written as e^(-1/x-1) / [2- e^(-1/x-1)]. How so?
1:34:31
The final answer shown in the video is equivalent to the form you've mentioned. To get your form, just multiply my final answer with e^((1/x)-1) to the numerator and denominator
@MathWorld-yp9od ah noted. Thank you soo much... your video really helped me in recapping chapters instantaneously ✨️
Thank you miss, do we have to show the working of the checking part ,does it carry working marks?
Hi, there are 2 methods to solve that question, i.e. graph and use definition of modulus function, then followed by squaring both sides to eliminate the square root. The checking part is to make sure the solutions satistify the inequality as there is only one modulus function in the inequality.
To me, the checking part is needed
At 7:10 why dont we see modulus of 14 as we saw for modulus of -2 ??
Modulus of 14 is 14. For your info, modulus value is always positive, so the modulus of negative value is positive. For the modulus of positive value, it is definitely a positive.
@@MathWorld-yp9od thank you so much
7:52 why does the mark scheme say WRITING AND for the inequalities is wrong ?
We always use OR and not AND, because as long as it satisfies 1 of the results,, then the inequality is valid. If you write AND, that means the inequality is only valid when satisfy both results
When you differentiate (ln t)² shouldn't it be 2/t
Where did ln t come from
d/dt(lnt) ^2=2(lnt)^1xd/dt(lnt)=(2lnt) /t according to the power rule or chain rule
Can you add the link to the solved paper in the description.
What link do you mean?
@@MathWorld-yp9od a link to the solved downloaded paper
you can save it to drive and give access to us
How do yoi know you need to use substitution in question9?
Use substitution method to solve the multiplication of same type of functions. In this case, sin and cos are both trigo and we cannot simplify them, hence use substitution method. U is the function with power but excluding the power
How is cot(pi/2) = 0?
Cot(pi/2)=1/tan(pi/2)=1/infinity =0
7b why include 90•?
Given 0°
Do you have notes?
Sorry, I don't have
hahaha, watching this an hour before the p3
Best of luck!
Sorry for qn 5 can't we use integration by parts
No, integration by parts is to integrate the product of 2 different types of functions. This is rational function involves polynomial in both numerator and denominator, hence, either ln, partial fractions or substitution method