Trig Integration (1 Viewer)

[thunderdax]

Evaluate ∫(cos²x/(1+sin²x))dx

Answer: -x+sqrt(2)tan^-1(sqrt(2)tanx)+C

I can get the -x bit but i just don't know how the rest of the integral works out. Need help please.

 

[shafqat]

Evaluate ∫(cos²x/(1+sin²x))dx

Answer: -x+sqrt(2)tan^-1(sqrt(2)tanx)+C

I can get the -x bit but i just don't know how the rest of the integral works out. Need help please.

I'm guessing you're stuck on the ∫1/((1+sin²x)) part left after the juggling with the cos^2x?
Divide top and bottom by cos^2 x, then use the pythagorean identities to simplify the denominator. Then use the substitution u = tanx.

 

[KFunk]

If you use the substitution u=tanx as shafqat suggested then you should end up with:

∫ 1/(1/2 + u²) - 1/(1+u²) dx

which gives you:

(√2)tan^-1√2.u - tan^-1u so when you sub in u = tanx you get your answer:

(√2)tan^-1(√2)tanx - tan^-1tanx +c
= -x + (√2)tan^-1(√2)tanx +c

 

[thunderdax]

Got it! Thanks for that.

 

[thunderdax]

Damm these trig integrals are annoying. Heres another one:
∫sqrt(1+sin2x)dx

 

[FinalFantasy]

Damm these trig integrals are annoying. Heres another one:
∫sqrt(1+sin2x)dx

∫sqrt(1+sin2x)dx=∫sqrt(sin²+2sinxcosx+cos²x)dx=∫sqrt(sinx+cosx)²dx
=∫(sinx+cosx)dx=-cosx+sinx+C