Evaluating a Definite Integral?

Question

The integral is from 0 to pi/2, and the problem is:

[sin(2x)] / [a+sin²(x)] dx

The problem also says:
Recall that sin(2x) = sin(x)cos(x), but I thought it would equal 2sin(x)cos(x)?

I have no idea how to begin this problem. Once I get that figured out, the rest should fall into place.

Thanks!

Answer 1

You are right; sin(2x)=2sin(x)cos(x)!

Let u=a+sin^2(x)

Then du=2sin(x)cos(x) dx=sin(2x) dx

Now do the substitution :)

int 1/u du = ln u + C = ln(a+sin^2(x)) + C

>> Plug in pi/2 subtract and plug in 0.

Answer 2

First off, sin 2x does indeed equal 2(sin x)(cos x). But instead of that, try using the Pythagorean identity to change the denominator to a + 1 - cos² x, and use another identity to replace cos² x with...something, I forget what.

HTH!

Answer 3

You are right, sin(2x) = 2 sin(x) cos(x)
solution: let a + [sin(x)]^2=u then du = 2 sin(x) cos(x)
supersede:
int { 2 sin(x) cos(x) / a + [sin(x)]^2 }
int { du / u } = ln( u ) = ln(a + sin(x)^2) + c
which c is a constant

Answer 4

You are right with your formula

2sin(x) * cos(x) /[a+sin^2(x)] dx

I woul make the change sin (x) = z so cos (x) dx =dz The integral becomes

2z/(a+z^2)* dz wich gives ln I a+z^2I = ln I a+sin^2(x)I beween pi/2 and 0

= ln Ia+1I - lnIaI = ln [I(a+1)/aI]