Question

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

Answer 1

$\frac{2x+2}{\sqrt{4x^2+8x+3}}=\frac{2(x+1)}{\sqrt{4\left(x^2\right(4+1)+3^2}}$

$=\frac{2(x+1)}{\sqrt{4(x+1{)}^2+3^2}}$

set $x+1=\frac{3}{2}\tan \theta$

$=\frac{2\cdot \frac{3}{2}\tan \theta }{\sqrt{4\times \frac{9}{4}{\tan }^2\theta +3^2}}=\frac{3\tan \theta }{\sqrt{3^2(1+{\tan }^2\theta )}}$

$=\frac{3\tan \theta }{3\sin \theta }=\frac{\sin \theta }{\cos \theta }\cdot \cos \theta =\sin \theta$

$={\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+\theta x+13}}\right)={\sin }^{-1}\left(\sin \theta \right)$

$=\theta$

$x+1=\frac{3}{2}\tan \theta$

$dx=\frac{3}{2}{\sin }^2\theta d\theta$

$\tan \theta =\frac{2}{3}(x+1)$

$\sin \theta =\sqrt{1+{\tan }^2\theta }$

$=\sqrt{1+\frac{4}{9}(x+1{)}^2}$

$=\frac{1}{3}\sqrt{4x^2+8x+13}$

$I=\int \theta \cdot \frac{3}{2}{\sin }^2\theta \cdot d\theta$

$=\frac{3}{2}\int \theta \cdot {\sec }^2\theta \cdot d\theta$

$=\frac{3}{2}[\theta \cdot \tan \theta -\int \tan \theta d\theta ]$

$=\frac{3}{2}[\theta \tan \theta -ln\sec \theta ]+c$

$=\frac{3}{2}\left[\frac{2}{3}\right(x+1\left)\tan \right\{\frac{2}{3}(x+1)\}$ $-ln\left(\frac{\sqrt{4n^2+dn+13}}{3}\right)+c$

$=\frac{3}{2}\cdot \frac{2}{3}(n+1){\tan }^{-1}\left\{\frac{2}{3}\right(n+1\left)\right\}-\frac{3}{2}[ln\sqrt{4n^2+dn+13}-ln3]+c$

$=(x+1){\tan }^{-1}\left\{\frac{2}{3}\right(n+1\left)\right\}-\frac{3}{2}\cdot \frac{1}{2}ln(4n^2+dn+13)+\frac{3}{2}ln3+c$ New laws

$=(x+1){\tan }^{-1}\left\{\frac{2}{3}\right(x+1\left)\right\}-\frac{3}{4}ln(4n^2+dn+13)+c^2$.