# Thread: Another one to integrate

1. ## Another one to integrate

1.)

2.)

Thanks a lot!!!!

2. Here's the first one

1.)

2.)

Thanks a lot!!!!
um, for the second one, i think you gave the answer key twice

4. ## sory for the typo

The question is
integral of sin(x)*(sin(cosx))dx

5. Originally Posted by Krizalid
Here's the first one

that's nice. i didn't even consider substitution. i ended up trying trig substitution, which was hell. but things seemed to be working out nicely until i ended up with x^4/4(x^2 + 4)^2 which is wrong.

The question is
integral of sin(x)*(sin(cosx))dx
This is a really easy one

Just set u = cos x and du = - sin x dx, therefore the integral becomes to -sin udu, which yields cos u + c and we've got that cos (cos x) + c.

The question is
integral of sin(x)*(sin(cosx))dx
we proceed by substitution:

let u = cos(x)
=> du = -sin(x) dx
=> - du = sin(x) dx
so our integral becomes:

- int{sin(u)}du = cos(u) + C = cos(cos(x)) + C

too late