int(tan(arcsin(x))) dx x=0,1 

That contained a bit of everything...trig substitution, u substitution, improper integral. Almost like an entire calculus class all at once

This is a good problem, it touches almost every calculus topic.

Tremendous video! As always.

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.

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.

thank you very much teacher.

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

May Allah grant me such knowledge.

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),

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)

Ill guess 2. I could be wrong because i tried it in my head.

sir please give integration of hyperbolic functions

1

I think you forget some think which is x=cos(thita) dx=-sin(thita) (d. thita) What do you think?

sir I have sent you a problem on your mail. Please solve that

Sin2x = 2tanx / 1+tan²x .replace x =sin⁻¹x

JEE aspirants dying of laughter

The title and the question dont match