Question

The value of integration $\int (\sin x{)}^{\ln x}\left\{\frac{\left(x\ln x\right)\cot x+\ln \sin x}{x}\right\}dx$ is

• A. $(\sin x{)}^{2\ln x}+c$
• B. $(\sin x{)}^{\ln x}+c$
• C. $\sqrt{(\sin x{)}^{\ln x}}+c$
• D. $\sin x+c$

Answer 1

The correct option is B $(\sin x{)}^{\ln x}+c$

Given :
$\int (\sin x{)}^{\ln x}\left\{\frac{\left(x\ln x\right)\cot x+\ln \sin x}{x}\right\}dx$
Let $t=(\sin x{)}^{\ln x}$
Applying $\ln$ on both sides,
$\ln t=\ln (\sin x{)}^{\ln x}$
$\Rightarrow \ln t=\ln x\cdot \ln \left(\sin x\right)$
differentiating w.r.t. $x$
$\frac{1}{t}\frac{dt}{dx}=\ln \sin x\cdot \left(\frac{1}{x}\right)+\ln x\cdot \left(\frac{1}{\sin x}\right)\cdot \cos x$
$=\frac{\ln \sin x}{x}+\ln x\cdot \cot x$
$\Rightarrow \frac{dt}{dx}=t\left\{\frac{\ln \sin x+\left(x\ln x\right)\cot x}{x}\right\}$
$\therefore dt=(\sin x{)}^{\ln x}\left\{\frac{\left(x\ln x\right)\cot x+\ln \sin x}{x}\right\}dx$
Now,
$\int (\sin x{)}^{\ln x}\left\{\frac{\left(x\ln x\right)\cot x+\ln \sin x}{x}\right\}dx$
$=\int 1dt$
$=t+c$
$=(\sin x{)}^{\ln x}+c$