# U-Substitution

• May 13th 2010, 03:48 PM
tbenne3
U-Substitution
I need help with the u-substituion. The first one is....

(2cosx)/(1-cos^2x)
• May 13th 2010, 03:49 PM
Chris L T521
Quote:

Originally Posted by tbenne3
I need help with the u-substituion. The first one is....

(2cosx)/(1-cos^2x)

Hint: $1-\cos^2x=\sin^2x$. Now what is the substitution?
• May 13th 2010, 03:57 PM
tbenne3
Quote:

Originally Posted by Chris L T521
Hint: $1-\cos^2x=\sin^2x$. Now what is the substitution?

I really don't know..
• May 13th 2010, 04:09 PM
skeeter
Quote:

Originally Posted by tbenne3
I really don't know..

better brush up on your basic identities ...

$\frac{2\cos{x}}{1 - \cos^2{x}} = \frac{2\cos{x}}{\sin^2{x}} = 2\csc{x}\cot{x}
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• May 13th 2010, 04:14 PM
tbenne3
Quote:

Originally Posted by skeeter
better brush up on your basic identities ...

$\frac{2\cos{x}}{1 - \cos^2{x}} = \frac{2\cos{x}}{\sin^2{x}} = 2\csc{x}\cot{x}
$

k thanks
• May 13th 2010, 05:04 PM
Prove It
Notice that if $u = \sin{x}$ then $du = \cos{x}\,dx$.

So $\int{\frac{2\cos{x}}{\sin^2{x}}\,dx} = 2\int{u^{-2}\,du}$

$= -2u^{-1} + C$

$= -\frac{2}{\sin{x}}+ C$

$= -2\csc{x} + C$.
• May 14th 2010, 04:16 AM
TheCoffeeMachine
You can also let $u = 1-\cos^2{x}$, then $\dfrac{du}{dx} = -2\cos{x}\sin{x} \Rightarrow {dx} = \dfrac{du}{-2\cos{x}\sin{x}}
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so then $\int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \int\dfrac{2\cos{x}}{u\left(-2\cos{x}\sin{x}\right)}\;{du} = -\int\dfrac{1}{u\sin{x}}\;{du}$. Now since $u = 1-\cos^2{x} = \sin^2{x}$, then $\sin{x} = \sqrt{u}$. So we have: $-\int\dfrac{1}{u\sin{x}}\;{du}= -\int\dfrac{1}{u\sqrt{u}}\;{du} = -\int\dfrac{1}{\sqrt{u^3}}\;{du} = \dfrac{-2}{\sqrt{u}}+C$ $\Rightarrow \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \dfrac{-2}{\sqrt{1-\cos^2{x}}}+C.
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