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

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

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

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

Because we know that $\displaystyle \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...

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

Re: Trigonometric Integrals Help

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

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

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

So

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

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

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

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

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

So

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

Ah yes, that does work!

But I'd do the proof like this.

You have

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

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

Then

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

Re: Trigonometric Integrals Help

Quote:

Originally Posted by

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

But I'd do the proof like this.

You have

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

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

Then

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

Thank you ILikeSerena. Your method is neater than mine.