Question

Evaluate $\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$

• A. $(x+1){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log (4x^2+8x+13)+c$
• B. $(x+1){\tan }^{-1}\left(\frac{2}{3}\right)-\frac{3}{4}\log (x^2+8x+13)+c$
• C. ${\tan }^{-1}\left(\frac{2x+2}{3}\right)+\frac{3}{4}\log (4x^2+8x+13)+c$
• D. none of these

Answer 1

The correct option is A $(x+1){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log (4x^2+8x+13)+c$

$I=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$
$=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{(2x+2{)}^2+3^2}}\right)dx$

Put $2x+2=3\tan \theta \Rightarrow 2dx=3{\sec }^2\theta d\theta$
$I=\int {\sin }^{-1}\left(\frac{3\tan \theta }{3\sec \theta }\right)\frac{3}{2}{\sec }^2\theta d\theta$
$=\frac{3}{2}\int \theta {\sec }^2\theta d\theta$
$=\frac{3}{2}\{\theta \tan \theta -\int \tan \theta d\theta \}$
$=\frac{3}{2}\{\theta \tan \theta -\log |\sec \theta |\}+c$
$=\frac{3}{2}\{\frac{2x+2}{3}{\tan }^{-1}\left(\frac{2x+2}{3}\right)-\log \left(\sqrt{1+{\left(\frac{2x+2}{3}\right)}^2}\right)\}+c$
$=\frac{3}{2}\left\{\frac{2}{3}\right(x+1){\tan }^{-1}\left(\frac{2}{3}\right(x+1\left)\right)-\log \sqrt{4x^2+8x+13}\}+c$
$\Rightarrow I=(x+1){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log (4x^2+8x+13)+c$