Question

Let $\int sin\left(2x\right)ln\left(cosx\right)dx=f\left(x\right){\left(cos\right)}^{2}x+C$ (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

Answer: 4

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