Integration Problem

**PMoNEY23** · April 14th 2009, 02:26 AM

antiderivative of [(sinx)^2 - (cosx)^2] / [sinx] dx

Do I start by splitting up? use u-sub situation?

**mr fantastic** · April 14th 2009, 02:34 AM

Originally Posted by PMoNEY23

antiderivative of [(sinx)^2 - (cosx)^2] / [sinx] dx

Do I start by splitting up? use u-sub situation?

You have $\int \! \frac{1 - 2 \cos^2 x}{\sin x} \, dx$ .

Substitute $u = \cos x \Rightarrow dx = - \frac{du}{\sin x}$ .

So you get $\int \! \frac{1 - 2 u^2}{\sin^2 x} \, (-du) = \int \! \frac{2 u^2 - 1}{1 - \cos^2 x} \, du = \int \! \frac{2 u^2 - 1}{1 - u^2} \, du = \int \! - 2 + \frac{1}{1 - u^2} \, du$ .

**Chop Suey** · April 14th 2009, 02:45 AM

Originally Posted by PMoNEY23

$\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx$

Do I start by splitting up? use u-sub situation?

Yes, you can do that.

$\begin{aligned} \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx &=& \int \frac{\sin^2{x}}{\sin{x}}~dx-\int \frac{\cos^2{x}}{\sin{x}}~dx \\ &=& \int \sin{x}~dx + \int -\cot{x}\csc{x}~dx \end{aligned}$

Which are 2 straightforward integrals.

**PMoNEY23** · April 14th 2009, 02:48 AM

I am up to the part where I have to integrate cotxcscx
but with u-substitution I don't get the correct answer
so let u=cscx and du=? then what?

**mr fantastic** · April 14th 2009, 02:49 AM

Originally Posted by Chop Suey

Yes, you can do that.

$\begin{aligned} \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx &=& \int \frac{\sin^2{x}}{\sin{x}}~dx-\int \frac{\cos^2{x}}{\sin{x}}~dx \\ &=& \int \sin{x}~dx + \int -\cot{x}\csc{x}~dx \end{aligned}$

Which are 2 straightforward integrals.

Provided you remember what the derivative of $\text{cosec} \, x$ is, of course.

**Chop Suey** · April 14th 2009, 02:54 AM

Originally Posted by mr fantastic

Provided you remember what the derivative of $\text{cosec} \, x$ is, of course.

Who doesn't?

EDIT:
@PMoney: Recall that $\frac{d}{dx}\csc{x} = -\csc{x}\cot{x}$

**Showcase_22** · April 14th 2009, 03:40 AM

$ \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx$ .

Isn't that a bit easier?

**mr fantastic** · April 14th 2009, 05:34 AM

Originally Posted by Showcase_22

$ \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx$ .

Isn't that a bit easier?

How?

**Showcase_22** · April 14th 2009, 11:01 AM

$ \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx $

$u=\frac{1}{sin \ x} \Rightarrow \frac{du}{dx}=\frac{-cos \ x}{sin^2 \ x}$

$\frac{dv}{dx}=cos \ 2x \Rightarrow v= \frac{1}{2}sin \ 2x$

$-\int \frac{cos(2x)}{sin \ x} \ dx= \frac{sin \ 2x}{2sin \ x}+\frac{1}{2}\int\frac{cos \ x \ sin \ 2x}{sin^2 \ x} \ dx$

... I guess it isn't easier after all =S

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