Question

\[\int\left( e^\text{log  x} + \sin x \right) \text{ cos x dx }\]

Answer 1

\[\int\left( e^\text{log  x} + \sin x \right) \text{ cos x dx }\]
\[ = \int \left( x + \sin x \right)\cos x dx \left( \because e^(log x = x \right)\]
\[ = \int \left( x \cos x + \sin x \cos x \right) dx\]
\[ = \int x \text{ cos x dx }+ \frac{1}{2}\int 2 \sin x \text{ cos x dx }\]
\[ = \int x_I \text{ cos}_{II} \text{    x  dx }+ \frac{1}{2} \int\text{ sin 2x dx }\]
\[ = \left[ x\int\text{ cos x dx }- \int\left\{ \frac{d}{dx}\left( x \right)\int\text
{ cos  x  dx } \right\}dx \right] + \frac{1}{2} \int\text{ sin  2x  dx }\]
\[ = x \sin x - \int1 . \text{ sin  x  dx} + \frac{1}{2}\left[ \frac{- \cos 2x}{2} \right] + C\]
\[ = x \sin x - \left( - \cos x \right) - \frac{1}{4}\cos 2x + C\]
\[ = x \sin x + \cos x - \frac{1}{4}\left( 1 - 2 \sin^2 x \right) + C\]
\[ = x \sin x + \cos x + \frac{\sin^2 x}{2} - \frac{1}{4} + C\]
\[ = x \sin x + \cos x + \frac{\sin^2 x}{2} + C'   \text{  where C' = C }- \frac{1}{4}\]