Trigonometric Integrals Help

Can you help me solve the integral of sin(x)/[cos^3 (x)] dx in three different ways?

So far I have:a.) tanx sec^2x dx

let u = tanx

du = sec^2x dx

= integral of u du

= (u^2)/2

= 1/2 tan^2 (x) + C

b.) sinx/cos^3x dx

(sinx)(cosx)^-3 dx

let u = cos x

du = -sinx

= - integral of u^-3

= 1/2 (cosx)^-2

= 1/2 sec^2x + C

I can't think of another way of solving it. Thanks in advance!

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

**vanillachyrae** Can you help me solve the integral of sin(x)/[cos^3 (x)] dx in three different ways?

So far I have:a.) tanx sec^2x dx

let u = tanx

du = sec^2x dx

= integral of u du

= (u^2)/2

= 1/2 tan^2 (x) + C

b.) sinx/cos^3x dx

(sinx)(cosx)^-3 dx

let u = cos x

du = -sinx

= - integral of u^-3

= 1/2 (cosx)^-2

= 1/2 sec^2x + C

I can't think of another way of solving it. Thanks in advance!

Hi vanillachyrae! :)

How about u = sec x?

Re: Trigonometric Integrals Help

Also let and then using integration by parts you can find the answer.

Because we know that

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

**x3bnm** Also let

and

then using integration by parts you can find the answer.

Because we know that

Nice try... but...

Re: Trigonometric Integrals Help

Sorry ILikeSerena. In my last post I accidentally typed instead of . I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

Let and

Now and

So

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

**x3bnm** Sorry ILikeSerena. In my last post I accidentally typed

instead of

. I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

Let

and

Now

and

So

Ah yes, that does work!

But I'd do the proof like this.

You have

Let

Then

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

**ILikeSerena** Ah yes, that does work!

But I'd do the proof like this.

You have

Let

Then

Thank you ILikeSerena. Your method is neater than mine.