# U-Substitution

#### tbenne3

I need help with the u-substituion. The first one is....

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

#### Chris L T521

MHF Hall of Fame
I need help with the u-substituion. The first one is....

(2cosx)/(1-cos^2x)
Hint: $$\displaystyle 1-\cos^2x=\sin^2x$$. Now what is the substitution?

#### tbenne3

Hint: $$\displaystyle 1-\cos^2x=\sin^2x$$. Now what is the substitution?
I really don't know..

#### skeeter

MHF Helper
I really don't know..
better brush up on your basic identities ...

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

#### tbenne3

better brush up on your basic identities ...

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

#### Prove It

MHF Helper
Notice that if $$\displaystyle u = \sin{x}$$ then $$\displaystyle du = \cos{x}\,dx$$.

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

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

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

$$\displaystyle = -2\csc{x} + C$$.

#### TheCoffeeMachine

You can also let $$\displaystyle u = 1-\cos^2{x}$$, then $$\displaystyle \dfrac{du}{dx} = -2\cos{x}\sin{x} \Rightarrow {dx} = \dfrac{du}{-2\cos{x}\sin{x}}$$ so then $$\displaystyle \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \int\dfrac{2\cos{x}}{u\left(-2\cos{x}\sin{x}\right)}\;{du} = -\int\dfrac{1}{u\sin{x}}\;{du}$$. Now since $$\displaystyle u = 1-\cos^2{x} = \sin^2{x}$$, then $$\displaystyle \sin{x} = \sqrt{u}$$. So we have: $$\displaystyle -\int\dfrac{1}{u\sin{x}}\;{du}= -\int\dfrac{1}{u\sqrt{u}}\;{du} = -\int\dfrac{1}{\sqrt{u^3}}\;{du} = \dfrac{-2}{\sqrt{u}}+C$$ $$\displaystyle \Rightarrow \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \dfrac{-2}{\sqrt{1-\cos^2{x}}}+C.$$