# Trig integrals

• Sep 8th 2008, 08:24 PM
symstar
Trig integrals
Two problems in this question, the first I was able to do but was getting an incorrect answer and the second I'm just plain stumped on.

FIRST Q:
$\displaystyle \int\cot^3{x}\csc^3{x}dx$

Here is what I've done so far... I'm coming close but my numbers aren't quite right.

$\displaystyle u=\csc^2{x}$
$\displaystyle du=-\cot{x}dx$
$\displaystyle \csc{x}=\sqrt{u}$

$\displaystyle \int (\csc^2{x}-1)\cot{x}\csc^3{x}dx$
$\displaystyle -\int (u-1)u\sqrt{u}du$
$\displaystyle -\int (u^{2}-u)u^{1/2}du$
$\displaystyle -\int u^{5/2}du + \int u^{3/2}du$
$\displaystyle \tfrac{2}{5}u^{5/2}-\tfrac{2}{7}u^{7/2}+C$

...and subbing in just leaves me with the wrong answer. Where did I go wrong?

SECOND Q:

$\displaystyle \int\csc{x}dx$

Thought to use $\displaystyle \csc{x}=\sqrt{1+\cot^2{x}}$ or $\displaystyle \frac{1}{\sin{x}}$, but neither really panned out (Worried)

• Sep 8th 2008, 08:29 PM
Chop Suey
Quote:

Originally Posted by symstar
$\displaystyle u=\csc^2{x}$
$\displaystyle du=-\cot{x}dx$
$\displaystyle \csc{x}=\sqrt{u}$

:confused: (Shake)(Shake)

The derivative of $\displaystyle \csc^2{x}$ is not $\displaystyle -\cot{x}$.

Pull one factor of $\displaystyle \csc{x}\cot{x}$ and recall that:

$\displaystyle (\csc{x})' = -\csc{x}\cot{x}$

$\displaystyle \cot^2{x} + 1 = \csc^2{x}$

As for your second question, multiply it with $\displaystyle \frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}}$ and see where that gets you.
• Sep 8th 2008, 08:34 PM
mr fantastic
Quote:

Originally Posted by symstar
[snip]
SECOND Q:

$\displaystyle \int\csc{x}dx$

Thought to use $\displaystyle \csc{x}=\sqrt{1+\cot^2{x}}$ or $\displaystyle \frac{1}{\sin{x}}$, but neither really panned out (Worried)