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Q. The value of definite integral $\int\limits_1^c\left(\frac{1}{x}-x+x \ln x\right) \sin x d x$ is equal to (where $e=\underset{x \rightarrow 0}{\text{Lim}} (1+x)^{\frac{1}{x}}$ )

Integrals

Solution:

image
$=((\sin x) \cdot \ln x)_1^c-\int\limits_1^e(\cos x) \cdot \ln x d x-((x \ln x-x) \cdot \cos x)_1^e+\int\limits_1^e(\ln x) \cdot(\cos x) d x$
$=(\sin e-\cos 1)$