## anonymous 5 years ago integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

1. anonymous

Recall sin 2x= 2[[(cos)^2]x]-1

2. anonymous

yup.. what nxt?

3. anonymous

4. anonymous

can you solve it out buddy.. i seem to get nowhere!

5. anonymous

You have to make an attempt. Don't get overwhelm. Your task is simple: choose a u. You could be right, you could be wrong. You won't harm anything.

6. anonymous

$\sin x+\cos x \over 9+16(2\cos ^{2}x-1)$ $\sin x+ \cos x \over 32\cos ^{2}x-7$ what do you think i should select as u

7. anonymous

We're not getting anywhere. I asked you twice: choose a u.

8. anonymous

i have no idea buddy.. i ll try later

9. watchmath

I am curious how we can really go from there. BTW $$\sin(2x)\neq 2\cos^2(x)-1$$. But $$\cos(2x)=2\cos^2(x)-1$$

10. anonymous

lol... i didnt evn realize that! rats!

11. anonymous

Good catch, watchmath. Apparently this question appears frequently in India exams. Solution depends on the relation 1 - sin 2x=(sinx - cos x)^2. Let sin x - cos x = t (cos x + sin x) dx = dt t^2= (sin x - cos x)^2 = 1 - sin 2x sin 2x = 1-t^2.$\int\limits_{0}^{\pi/2}(\sin x + \cos x)/(9+16 \sin 2x)$=$\int\limits_{0}^{\pi/2}dt/(9+16(1-t ^{2})$

12. watchmath