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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Recall sin 2x= 2[[(cos)^2]x]1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you solve it out buddy.. i seem to get nowhere!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin x+\cos x \over 9+16(2\cos ^{2}x1)\] \[\sin x+ \cos x \over 32\cos ^{2}x7\] what do you think i should select as u

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're not getting anywhere. I asked you twice: choose a u.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have no idea buddy.. i ll try later

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I am curious how we can really go from there. BTW \(\sin(2x)\neq 2\cos^2(x)1\). But \(\cos(2x)=2\cos^2(x)1\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol... i didnt evn realize that! rats!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good 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 = 1t^2.\[\int\limits_{0}^{\pi/2}(\sin x + \cos x)/(9+16 \sin 2x)\]=\[\int\limits_{0}^{\pi/2}dt/(9+16(1t ^{2})\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd3bbd699508b0bc2d6a5a4
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