Note: When you're simplifying trig(inverse trig(x)), be cautious of the domain, range, and sign! Whenever you pull out the right triangle you are implicitly assuming you're working in the 1st quadrant. And that's not always gonna be the case. For example, sin(arcsec(x)) should simplify to sqrt(x^2 - 1)/x, but that only works whenever arcsec(x) is in the FIRST quadrant, so 0
The simplest would be just a substitution of arcsin x to u. Let u = arcsin x sin u = x cos u du = dx Integrand will become tan u cos u => sin u Limits will be from π/2 to 0.
It is always worth trying to substitute a whole annoying function as you did or if ou have function composition the inside function as u. Works out very nice in this case
Since the sqrt(u) of u was in the numerator did you have to take the limit... That is, can you come up with an example where the integrand is undefined between the limit of integration, but the integral is defined over the limit of integration, but by direct substitution of these limits will give you the wrong answer? That would be quite interesting!
The function used is also known as the Einstein's Relativistic Function, as it calculates time dilation at speeds moving relative to c, the speed of light.
My solution to this problem: First, it's helpful to first simplify tan(arcsin(x)). tan = sin/cos, so tan(arcsin(x)) = x/cos(arcsin(x)). cos(arcsin(x)) = sqrt(1 - x²) is a known identity that follows from the pythagorean identity: cos(y)² + sin(y)² = 1 cos(y)² = 1 - sin(y)² cos(y) = sqrt(1 - sin(y)²) cos(arcsin(x)) = sqrt(1 - x²) Applying this identity yields I = int tan(arcsin(x)) dx = int x/sqrt(1-x²) dx. What should be noted is that a lot of what I did so far isn't valid in general (like taking sqrt on both sides, or saying sin(arcsin(x))=x), but holds true when x is in [0, 1], which is indeed granted in our case. To evaluate this integral, we can substitute u=1-x², du=-2x dx and get I = int x/sqrt(1-x²) dx = -1/2 int 1/sqrt(1-x²) (-2x dx) = -1/2 int 1/sqrt(u) du = -1/2 int u^(-1/2) du (use power rule to solve integral) = -u^(1/2) = -(1-x²)^(1/2) Now we just evaluate that at x=1 and x=0, and get: I = -(1-1²)^(1/2) - (-(1-0²)^(1/2)) = 1
![int[sin^4(x)]](http://i.ytimg.com/vi/Lwmn-ZTCBMI/mqdefault.jpg)
This is a good problem, it touches almost every calculus topic.
That contained a bit of everything...trig substitution, u substitution, improper integral. Almost like an entire calculus class all at once
Note: When you're simplifying trig(inverse trig(x)), be cautious of the domain, range, and sign!
Whenever you pull out the right triangle you are implicitly assuming you're working in the 1st quadrant. And that's not always gonna be the case.
For example, sin(arcsec(x)) should simplify to sqrt(x^2 - 1)/x, but that only works whenever arcsec(x) is in the FIRST quadrant, so 0
The simplest would be just a substitution of arcsin x to u.
Let u = arcsin x
sin u = x
cos u du = dx
Integrand will become tan u cos u => sin u
Limits will be from π/2 to 0.
That is what Newton did. He called it theta instead of u.
Yes, this is how I did it as well.
It is always worth trying to substitute a whole annoying function as you did or if ou have function composition the inside function as u.
Works out very nice in this case
Tremendous video! As always.
Since the sqrt(u) of u was in the numerator did you have to take the limit... That is, can you come up with an example where the integrand is undefined between the limit of integration, but the integral is defined over the limit of integration, but by direct substitution of these limits will give you the wrong answer? That would be quite interesting!
May Allah grant me such knowledge.
The function used is also known as the Einstein's Relativistic Function, as it calculates time dilation at speeds moving relative to c, the speed of light.
My solution to this problem:
First, it's helpful to first simplify tan(arcsin(x)).
tan = sin/cos, so tan(arcsin(x)) = x/cos(arcsin(x)).
cos(arcsin(x)) = sqrt(1 - x²) is a known identity that follows from the pythagorean identity:
cos(y)² + sin(y)² = 1
cos(y)² = 1 - sin(y)²
cos(y) = sqrt(1 - sin(y)²)
cos(arcsin(x)) = sqrt(1 - x²)
Applying this identity yields I = int tan(arcsin(x)) dx = int x/sqrt(1-x²) dx.
What should be noted is that a lot of what I did so far isn't valid in general (like taking sqrt on both sides, or saying sin(arcsin(x))=x), but holds true when x is in [0, 1], which is indeed granted in our case.
To evaluate this integral, we can substitute
u=1-x², du=-2x dx
and get
I = int x/sqrt(1-x²) dx
= -1/2 int 1/sqrt(1-x²) (-2x dx)
= -1/2 int 1/sqrt(u) du
= -1/2 int u^(-1/2) du (use power rule to solve integral)
= -u^(1/2)
= -(1-x²)^(1/2)
Now we just evaluate that at x=1 and x=0, and get:
I = -(1-1²)^(1/2) - (-(1-0²)^(1/2))
= 1
thank you very much teacher.
evaluate the integral of I = ∫[1,0] (x + y) dx from point A(0,1) to point B(0,-1) along the semicircle y = √(1-x²),
Ill guess 2. I could be wrong because i tried it in my head.
Im wrong. Disregard.
tan(arcsin(x)) = sin(arcsin(x))/cos(arcsin(x))
tan(arcsin(x)) = x/sqrt(1-(sin(arcsin(x)))^2)
tan(arcsin(x)) = x/sqrt(1-x^2)
Also known as the Einstein's Relativistic Function, as it calculates time dilation at speeds moving relative to c, the speed of light.
sir please give integration of hyperbolic functions
I think you forget some think which is
x=cos(thita)
dx=-sin(thita) (d. thita)
What do you think?
Sin2x = 2tanx / 1+tan²x .replace x =sin⁻¹x
1
sir I have sent you a problem on your mail. Please solve that
JEE aspirants dying of laughter
The title and the question dont match
They do wdym?
Inverse sin is also called arcsin
@@J.B.L2227 it was arctan before