Question

$\int e^x(\log \sin x+cose{c}^{2}x)dx$ is equal to

• A. $e^x\log \left(\sin x\right)+c$
• B. $e^x\left\{\log \right(\sin x)-\cot x\}+c$
• C. $e^x\left\{\log \right(\sin x)+\cot x\}+c$
• D. $e^x\cot x+c$

Answer 1

Answer: (B)

$e^x\left\{\log \right(\sin x)-\cot x\}+c$

$\int e^x(\log \sin x+cose{c}^{2}x)dx$

$=\int e^x\left(\log \sin x\right)dx+\int (cose{c}^{2}x{)}^{ex}dx.$

Integrating by parts,

$\log \left(\sin x\right)\int e^x-\int \frac{\cot x}{\sin x}{e}^{x}dx+\int \left(cose{c}^{2}x\right)e^{x}dx$

$=\log \left(\sin x\right)e^x-\int \cot x{e}^x\cdot dx+\int \left(cose{c}^{2}x\right)e^{x}dx$

$=\log \left(\sin x\right)e^x-[\cot x{e}^x+\int cose{c}^{2}x{e}^{x}dx]+\int \left(cose{c}^{2}x\right)e^{x}dx$

$=e^x\left[\log \right(\sin x)-\cot x]+c$