Question

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

Answer 1

Let $ I=\int \limits sin ^{-1} \bigg (\frac {2x+2}{\sqrt {4x^2+8x+13 }}\bigg )dx$
$ = \int \limits sin ^{-1}\bigg (\frac {2x+2}{\sqrt {(2x+2^2)+9}}\bigg )dx$
Put 2x+2$=3 tan \, heta\Rightarrow 2dx=3 \, sec^2 heta d heta$
$\therefore I= \int \limits sin ^{-1} \bigg (\frac {3tan heta}{{\sqrt {9tan^2 heta+9}}}\bigg ). \frac {3}{2}sec^2 heta d heta$
$= \int \limits sin ^{-1} \bigg (\frac {3tan \, heta}{3sec heta}\bigg ). \frac {3}{2}sec^2 heta\, d heta$
$= \int \limits sin^{-1}\bigg (\frac {sin \, heta}{cos \, heta.sec \, heta}\bigg ). \frac {3}{2}sec^2 heta \, d heta$
$ = \frac {3}{2} \int \limits sin^{-1} (sin \, heta).sec^2 heta\, d heta $
$ = \frac {3}{2} \int \limits heta.sec^2 heta \, d heta= \frac {3}{2}[ heta. tan\, heta- \int \limits 1. tan \, heta d heta] $
$ = \frac {3}{2}[ heta\, tan \, heta-log \, sec \, heta]+c $
$ = \frac {3}{2}[tan^{-1} \bigg (\frac {2x+2}{3}\bigg ). \bigg (\frac {2x+2}{3} \bigg )$
$ -log \sqrt {1+ \bigg (\frac {2x+2}{3}\bigg )^2}\bigg ]+ c_1$
$ = (x+1) tan^{-1}\bigg (\frac {2x+2}{3}\bigg )- \frac {3}{4}log \bigg [1+ \bigg (\frac {2x+2}{3}\bigg )^2\bigg ]+c_1$
$ = (x+1) tan^{-1} \bigg (\frac {2x+2}{3}\bigg )- \frac {3}{4} log (4x^2+8x+13)+c$
$ \Bigg [ let \, \frac{3}{2} log\, 3 + c_1 = c \Bigg ]$