Answer to Question #194456 in Calculus for she

Question

1/x( √4+x^2)^3 using substitution x = 2 tan ϴ

Answer 1

Substitution

$$x=2\tan \theta$$

$$dx=\dfrac{2d\theta}{\cos^2\theta }$$

$$\sqrt{4+x^2 }=\sqrt{4+(2\tan \theta)^2 }=\dfrac{2}{\cos\theta }$$

$$\int\dfrac{dx}{x(\sqrt{4+x^2})^3}=\int\dfrac{2d\theta}{\cos^2\theta (2\tan \theta )(\dfrac{2}{\cos\theta })^3}$$

$$=\int\dfrac{d\theta}{8\sin \theta }$$

$\cos \theta=u$

$$du=-\sin \theta d\theta$$

$$\sin^2 \theta =1-u^2$$

$$\int\dfrac{d\theta}{8\sin \theta }=-\int\dfrac{ du}{8(1-u^2)}$$

$$=-\int\dfrac{ du}{16(1-u)}-\int\dfrac{ du}{16(1+u)}$$

$$=\dfrac{1}{16}\ln|1-u|-\dfrac{1}{16}\ln|1+u|+C$$

$$=\dfrac{1}{16}\ln(1-\cos\theta)-\dfrac{1}{16}\ln(1+\cos\theta)+C$$

$$=\dfrac{1}{16}\ln(1-\cos(\arctan\dfrac{x}{2}))$$

$$-\dfrac{1}{16}\ln(1+\cos(\arctan\dfrac{x}{2}))+C$$