Question

Let $\displaystyle \int sin\left(2 x\right)ln\left(cos x\right)dx=f\left(x\right)\left(cos\right)^{2}x+C$ (where, $C$ is the constant of integration) and $f\left(0\right)=\frac{1}{2}$ . If $f\left(\frac{\pi }{3}\right)$ is equal to $\frac{1}{a}+lnb$ , then the value of $a+b$ is

Answer 1

Given integral is
$I=\displaystyle \int 2sinxcosxln\left(cos x\right)dx$
Let, $cosx=t$
$\Rightarrow -sinxdx=dt$
$\Rightarrow I=-2\displaystyle \int tlntdt$
Using Integration by parts
$I=-2\left(\frac{t^{2}}{2} ln t - \displaystyle \int \frac{t^{2}}{2} \frac{1}{t} d t\right)$
$I=-t^{2}lnt+\frac{t^{2}}{2}+C$
$I=\left(cos\right)^{2}x\left(\frac{1}{2} - ln cos x\right)+C$
$f\left(x\right)=\frac{1}{2}-ln\left(cos x\right)$
$f\left(\frac{\pi }{3}\right)=\frac{1}{2}-ln\left(\frac{1}{2}\right)$
$=\frac{1}{2}+ln2$
$\Rightarrow a=b=2$
$\therefore a+b=4$