Question

integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

Answer 1

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

Answer 2

yup.. what nxt?

Answer 3

Choose your u.

Answer 4

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

Answer 5

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.

Answer 6

\[\sin x+\cos x \over 9+16(2\cos ^{2}x-1)\]
\[\sin x+ \cos x \over 32\cos ^{2}x-7\]

what do you think i should select as u

Answer 7

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

Answer 8

i have no idea buddy.. i ll try later

Answer 9

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

Answer 10

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

Answer 11

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.\[\int\limits_{0}^{\pi/2}(\sin x + \cos x)/(9+16 \sin 2x)\]=\[\int\limits_{0}^{\pi/2}dt/(9+16(1-t ^{2})\]

Answer 12

http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd3bbd699508b0bc2d6a5a4