Question

integrate 1/(sinx+cosx)^2

Answer 1

everyone hates integration lol....

Answer 2

Have you tried expanding out the denominator? It seems like we can get some trigonometric identities involved with sin^2 x + 2 sin x cos x + cos^2 x...

Answer 3

\[\int\limits_{}^{}\frac{ 1 }{ (\sin x + \cos x)^2 }\]

Answer 4

I would expand the denominator as well. U-substitution wouldn't help in this form

Answer 5

i did, actually

Answer 6

i tried sub, integration by parts, and whatever identities i could put in there, still no luck... :(

Answer 7

expand denominator and get
sin^2(x) + cos^2(x) + 2sinxcosx
sin^2 + cos^2 = 1
and 2sinxcosx = sin(2x)

so your integrating
1/ 1 + sin(2x)

Answer 8

\[\int\limits\limits_{}^{}\frac{ 1 }{ \sin^2x+2\cos x \sin x + \cos^2x }\]\[\int\limits\limits\limits_{}^{}\frac{ 1 }{ 2\cos x \sin x+(\sin^2x+ \cos^2x) }\]

Answer 9

i tried all of the above, still counldn't see a way...

Answer 10

\[\int\limits\limits\limits\limits_{}^{}\frac{ 1 }{ 2\cos x \sin x+1 }\]

Answer 11

nothing i haven't tried so far, eagerly waiting for the big moment ....

Answer 12

\( \displaystyle \int \frac{1}{1 + \sin 2x } \)
How about multiplying top and bottom by 1 - sin 2x? This way, you'll get 1 - sin^2 2x on the bottom, which simplifies to cos^2 2x, and a mere 1 - sin 2x on top. Then we can break the fraction up...

Answer 13

o, u r right :D, got it thx

Answer 14

well, that was fun..... :D

Answer 15

You're welcome! :)

Answer 16

can't believe i didnt remember to try the conjugate lol, smh