Question

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

Answer 1

Let $I=\int si{n}^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$
$=\int si{n}^{-1}\left(\frac{2x+2}{\sqrt{(2x+2^2)+9}}\right)dx$
Put 2x+2$=3tanheta\Rightarrow 2dx=3se{c}^{2}hetadheta$
$\therefore I=\int si{n}^{-1}\left(\frac{3tanheta}{\sqrt{9ta{n}^{2}heta+9}}\right).\frac{3}{2}se{c}^{2}hetadheta$
$=\int si{n}^{-1}\left(\frac{3tanheta}{3secheta}\right).\frac{3}{2}se{c}^{2}hetadheta$
$=\int si{n}^{-1}\left(\frac{sinheta}{cosheta.secheta}\right).\frac{3}{2}se{c}^{2}hetadheta$
$=\frac{3}{2}\int si{n}^{-1}(sinheta).se{c}^{2}hetadheta$
$=\frac{3}{2}\int heta.se{c}^{2}hetadheta=\frac{3}{2}[heta.tanheta-\int 1.tanhetadheta]$
$=\frac{3}{2}[hetatanheta-logsecheta]+c$
$=\frac{3}{2}[ta{n}^{-1}(\frac{2x+2}{3}).(\frac{2x+2}{3})$
$-log\sqrt{1+(\frac{2x+2}{3}{)}^2}]+c_1$
$=(x+1)ta{n}^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}log[1+(\frac{2x+2}{3}{)}^2]+c_1$
$=(x+1)ta{n}^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}log(4x^2+8x+13)+c$
$[let\frac{3}{2}log3+c_1=c]$