S(0 to pi/2) cosx/(sinx)^1/2 dx ?

Question

tried substitution u=sinx but this cancelled out cosx which it seems like I needed for the derivitive of the inside. Another tricky integral...or are these considered simple by people who are good at calculus. The solution to this interval does not seem obvious to me. What insight am i missing I wonder?

Answer 1

Integral (0 to pi/2, cos(x)/(sin(x))^(1/2) dx )

We have to use substitution to solve this. But first, I'll present an intermediate step to show the substitution in action. Moving the cos(x) next to the dx, we get

Integral (0 to pi/2, 1/[sin(x)]^(1/2) cos(x) dx )

Here's where we use substitution. Our bounds will change also.

Let u = sin(x). (When x = 0, u = sin(0) = 0. When x = pi/2,
u = sin(pi/2) = 1)
du = cos(x) dx.

Note that our bounds of integration have changed from 0 to 1.
Also, cos(x) dx is the tail end of our integral, so du will be the tail end of our integral.

Integral (0 to 1, 1/u^(1/2) du )

Integral (0 to 1, u^(-1/2) du )

Use the reverse power rule.

2u^(1/2) {evaluated from 0 to 1}

[2(1)^(1/2) - 2(0)^(1/2)]

2(1) - 0

2 - 0

2

EDIT: Thanks Steiner!

Answer 2

Actually, it's simple. Just observe that cos(x) is the derivative of sin(x). If you put u = sin(x), then du = cos(x) dx and you get an integral of the type

Int u^(-1/2) du = 1/(-1/2 + 1) u^(-1/2 + 1) = 2 u^(1/2) + C

Since u = sin(x), we get [2 (sin(x))^(1/2)] (from 0 to pi/2] =

2 [sin(pi/2)^(1/2) - sin(0)^(1/2)] 2 [1 - 0] = 2.

The previous person made a small mistake at the end of his work.

Answer 3

∫cosec^2 (x - π/2)dx-------------(1) Solution : Let x - π/2 = y---------------(2) then, x = y + π/2 differentiating on both sides, dx = dy---------------(3) substituting (2) and (3) in (1), ∫cosec^2 (x - π/2)dx simplifies to ∫cosec^2 (y)dy we have the direct relation : ∫cosec^2 (y)dy = -coty + c therefore, ∫cosec^2 (y)dy = -coty + c from (2), we have : x - π/2 = y FINAL SOLUTION: ∫cosec^2 (x - π/2)dx = -cot(x - π/2) + c, where ''c'' is an arbitrary constant