Question

Find

$\int ta{n}^{-1}\frac{x}{\sqrt{a^2-x^2}}dx;\left|x\right|⟨a$

Answer 1

${\text{tan}}^{-1}\frac{x}{\sqrt{a^2-x^2}}$

$\sin \theta =\frac{x}{a}$

$⟹\cos \theta =\frac{\sqrt{a^2-x^2}}{a}⟹\tan \theta =\frac{x}{\sqrt{a^2-x^2}}⟹\theta ={\text{tan}}^{-1}\frac{x}{\sqrt{a^2-x^2}}$

$⟹{\text{tan}}^{-1}\frac{x}{\sqrt{a^2-x^2}}={\text{sin}}^{-1}\frac{x}{a}$

$\int {\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}dx=\int {\text{sin}}^{-1}\frac{x}{a}dx=x{\text{sin}}^{-1}\frac{x}{a}+2\sqrt{a^2-x^2}+c$