Trig integrals

**symstar** · September 8th 2008, 08:24 PM

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

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

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

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

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

SECOND Q:

$\int\csc{x}dx$

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

Thanks in advance.

**Chop Suey** · September 8th 2008, 08:29 PM

Originally Posted by symstar

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

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

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

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

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

As for your second question, multiply it with $\frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}}$ and see where that gets you.

**mr fantastic** · September 8th 2008, 08:34 PM

Originally Posted by symstar

[snip]
SECOND Q:

$\int\csc{x}dx$

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

Thanks in advance.

The conventional approach is to use the Weierstrass substitution: The Weierstrass Substitution Example

**symstar** · September 8th 2008, 08:57 PM

Can't believe I made that mistake... I was thinking of the integral and not derivative. My class isn't up to the Weierstrass substitution yet. Also, I find it funny that a proof for the integral of csc(x) I was viewing stated that the strategy was not obvious... oi!

Thanks for the help.

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