Trig...! :( (1 Viewer)

[Mikearun]

can anyone show me how to integrate cos^2x/sin^4x ? i dont know where to start

 

[Riviet]

cos²x/sin⁴x = cos²x/sin²x . 1/sin²x
=cot²x / sin²x
=cot²x.cosec²x dx
let u = cot x
du=-cosec²x dx

.'. ∫ cos²x/sin⁴x dx = -∫ -cot²x.cosec²x dx
=-∫ u² du
=-u³ / 3 + C
=-cot³x / 3 + C

 

[Mikearun]

so what do you normally start with for this type of question ?

 

[Riviet]

At first, I was kind of stuck just like you. With this type of trig integral I would try all soughts of manipulations, mainly using identities of 1+tan²x, 1+cot²x and all the various sine and cosine idenities to get the integral into a form that I could integrate, alot of the time it will be a substitution.

 

[Mikearun]

So u just basically have to try all possibilities to get the answer for this type , isnt it ?
I SEE NOW ,, thanks for help

 

[Riviet]

Try things out, experiment with different identities, do some more of these questions and you should get better at recognising what method to use or what substitution to make, it all comes together with practise. Glad to help.