Integrals — Class 12 Maths Solution

: Let $I = \int\limits_0^1 {\left[ {x{e^x} + \sin \left( {\cfrac{{\pi x}}{4}} \right)} \right]dx} = \int\limits_0^1 {x{e^x}dx} + \int\limits_0^1 {\sin \cfrac{{\pi x}}{4}dx}$

$= \left[ {x{e^x} - \int {\left( {\cfrac{d}{{dx}}\left( x \right) \cdot \int {{e^x}dx} } \right)dx} } \right]_0^1 - \left[ {\cfrac{{\cos \cfrac{{\pi x}}{4}}}{{\cfrac{\pi }{4}}}} \right]_0^1$

$= \left[ {x{e^x} - \int {{e^x}} dx} \right]_0^1 - \cfrac{4}{\pi }\left[ {\cos \cfrac{{\pi x}}{4}} \right]_0^1$

$= \left[ {x{e^x} - {e^x}} \right]_0^1 - \cfrac{4}{\pi }\left( {\cos \cfrac{\pi }{4} - \cos 0} \right)$

$= \left( {{e^1} - 0} \right) - \left( {{e^1} - {e^0}} \right) - \cfrac{4}{\pi }\left( {\cfrac{1}{{\sqrt 2 }} - 1} \right)$

$= e - e + 1 - \cfrac{4}{{\pi \sqrt 2 }} + \cfrac{4}{\pi } = 1 + \cfrac{4}{\pi } - \cfrac{{2\sqrt 2 }}{\pi }$