1. ## 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!

2. ## Re: Trigonometric Integrals Help

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?

3. ## Re: Trigonometric Integrals Help

Also let $u = \tan(x)$ and $dv = \sec^2(x)$ then using integration by parts you can find the answer.

Because we know that $\int \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx$

4. ## Re: Trigonometric Integrals Help

Originally Posted by x3bnm
Also let $u = \tan^2(x)$ and $dv = \sec^2(x)$ then using integration by parts you can find the answer.

Because we know that $\int \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx$
Nice try... but...

$udv = \tan^2 x \sec^2 x dx = \frac{\sin^2x}{\cos^4x} \ne \frac{\sin x}{\cos^3x}$

5. ## Re: Trigonometric Integrals Help

Sorry ILikeSerena. In my last post I accidentally typed $u = \tan^2(x)$ instead of $u = \tan(x)$. I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

Let $u = \tan(x)$ and $dv = \sec^2(x)\,\,dx$
Now $du = \sec^2(x)\,\,dx$ and $v = \tan(x)$

So
\begin{align*}\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx \\ =& \tan^2(x) - \frac{\tan^2(x)}{2} + C\\ =& \frac{\tan^2(x)}{2} + C \end{align*}

6. ## Re: Trigonometric Integrals Help

Originally Posted by x3bnm
Sorry ILikeSerena. In my last post I accidentally typed $u = \tan^2(x)$ instead of $u = \tan(x)$. I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

Let $u = \tan(x)$ and $dv = \sec^2(x)\,\,dx$
Now $du = \sec^2(x)\,\,dx$ and $v = \tan(x)$

So
\begin{align*}\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx \\ =& \tan^2(x) - \frac{\tan^2(x)}{2} + C\\ =& \frac{\tan^2(x)}{2} + C \end{align*}
Ah yes, that does work!
But I'd do the proof like this.

You have

$\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx$

Let $I =\int \tan(x) \sec^2(x)\,\, dx$
Then

\begin{align*}I =& \tan^2(x) - I \\ 2I =& \tan^2(x)\\ I =& \frac{\tan^2(x)}{2} + C \end{align*}

7. ## Re: Trigonometric Integrals Help

Originally Posted by ILikeSerena
Ah yes, that does work!
But I'd do the proof like this.

You have

$\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx$

Let $I =\int \tan(x) \sec^2(x)\,\, dx$
Then

\begin{align*}I =& \tan^2(x) - I \\ 2I =& \tan^2(x)\\ I =& \frac{\tan^2(x)}{2} + C \end{align*}
Thank you ILikeSerena. Your method is neater than mine.