Question

$\int {\sin }^{-1}\left\{\frac{2x+2}{\sqrt{4x^2+8x+13}}\right\}dx=\frac{k}{2}[\theta \tan \theta -\log \sec \theta ]$ where $\tan \theta =\frac{2x+2}{3}.$ Find the value of $k$.

Answer 1

Let $I=\int {\sin }^{-1}\left\{\frac{2x+2}{\sqrt{4x^2+8x+13}}\right\}dx$
Write $4x^2+8x+13={(2x+2)}^2+9$
Put $2x+2=t\Rightarrow dx=\frac{1}{2}dt$
Therefore
$I=\frac{1}{2}\int {\sin }^{-1}\frac{t}{\sqrt{t^2+9}}dt$
Now put $t=3\tan \theta \Rightarrow 3{\sec }^2\theta d\theta =dt$
Hence
$I=\frac{1}{2}\int {\sin }^{-1}\frac{3\tan \theta }{3\sec \theta }.3{\sec }^2\theta d\theta =\frac{3}{2}\int {\sec }^2\theta d\theta$
$=\frac{3}{2}[\theta \tan \theta -\int \tan \theta d\theta ]=\frac{3}{2}[\theta \tan \theta -\log \sec \theta ]$
Where $\tan \theta =\frac{t}{3}=\frac{2x+2}{3}$