# integration by substitution

#### wik_chick88

$$\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx$$

it says to substitute $$\displaystyle x = 2 tan \theta$$

i get down to $$\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta$$

no idea what to do from here!

#### Sudharaka

$$\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx$$

it says to substitute $$\displaystyle x = 2 tan \theta$$

i get down to $$\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta$$

no idea what to do from here!
Dear wik_chick,

$$\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta$$

$$\displaystyle \frac{1}{4}\int\frac{cos\theta}{sin^{2}\theta}~d\theta$$

$$\displaystyle \frac{1}{4}\int{cot\theta~{cosec\theta}~d\theta}$$

$$\displaystyle -\frac{1}{4}\int\frac{d}{d\theta}cosec\theta~{d\theta}$$

Hope you can continue from here.

#### Prove It

MHF Helper
$$\displaystyle \int \frac{1}{x^2 \sqrt {4+x^2}} dx$$

it says to substitute $$\displaystyle x = 2 tan \theta$$

i get down to $$\displaystyle \frac{1}{4} \int \frac{cos \theta}{1 - cos^2 \theta} d \theta$$

no idea what to do from here!
You should find that when you make the substitutions:

Remember that $$\displaystyle 1 - \cos^2{\theta} = \sin^2{\theta}$$.

Then the integral becomes

$$\displaystyle \frac{1}{4}\int{\frac{\cos{\theta}}{\sin^2{\theta}}\,d\theta}$$.

Now let $$\displaystyle u = \sin{\theta}$$ so that $$\displaystyle du = \cos{\theta}\,d\theta$$ and the integral becomes

$$\displaystyle \frac{1}{4}\int{\frac{1}{u^2}\,du}$$

$$\displaystyle = \frac{1}{4}\int{u^{-2}\,du}$$

$$\displaystyle = -\frac{1}{4}u^{-1} + C$$

$$\displaystyle = -\frac{1}{4u} + C$$

$$\displaystyle = -\frac{1}{4\sin{\theta}} + C$$.

Now remembering that $$\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$$

$$\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\sqrt{1 + \left(\frac{x}{2}\right)^2}}$$

$$\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\sqrt{\frac{4 + x^2}{4}}}$$

$$\displaystyle \sin{\theta} = \frac{\frac{x}{2}}{\frac{\sqrt{4 + x^2}}{2}}$$

$$\displaystyle \sin{\theta} = \frac{x}{\sqrt{4 + x^2}}$$.

Plugging this back into the solution gives:

$$\displaystyle -\frac{1}{4\sin{\theta}} + C = -\frac{1}{\frac{4x}{\sqrt{4 + x^2}}} + C$$

$$\displaystyle = -\frac{\sqrt{4 + x^2}}{4x} + C$$.

wik_chick88