Question

The value of definite integral $\underset{1}{\overset{c}{\int }}\left(\frac{1}{x}-x+x\ln x\right)\sin xdx$ is equal to (where $e=\underset{x\to 0}{\mathrm{Lim}}(1+x{)}^{\frac{1}{x}}$ )

• A. $\sin e-\cos 1$
• B. $\cos e-\sin 1$
• C. $\sin e+\cos 1$
• D. $\cos e+\sin 1$

Answer 1

Answer: (A)

$=((\sin x)\cdot \ln x{)}_1^c-\underset{1}{\overset{e}{\int }}(\cos x)\cdot \ln xdx-((x\ln x-x)\cdot \cos x{)}_1^e+\underset{1}{\overset{e}{\int }}(\ln x)\cdot (\cos x)dx$
$=(\sin e-\cos 1)$