1. ## Integration Problem

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

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

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

3. 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.

4. 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?

5. 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.

6. 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 $\frac{d}{dx}\csc{x} = -\csc{x}\cot{x}$

7. $
\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?

8. 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?

9. $
\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