Question

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

Answer 1

Consider

\[f\left( x \right) = 2\sin\left| x \right| + \cos\left| x \right|\]

Now,

\[f\left( - x \right) = 2\sin\left| - x \right| + \cos\left| - x \right| = 2\sin\left| x \right| + \cos\left| x \right| = f\left( x \right)\]

\[\therefore \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]
\[ = 2 \int_0^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx ................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]
\[ = 2 \int_0^\frac{\pi}{2} \left( 2\sin x + \cos x \right)dx ......................\left[ \left| x \right| = \begin{cases}x, & \text{if }x \geq 0 \\ - x, & \text{if }x < 0\end{cases} \right]\]
\[ = 4 \int_0^\frac{\pi}{2} \sin x\ dx + 2 \int_0^\frac{\pi}{2} \cos x\ dx\]

\[= \left.4 \times \left( - \cos x \right)\right|_0^\frac{\pi}{2} + \left.2 \times \sin x\right|_0^\frac{\pi}{2} \]
\[ = - 4\left( \cos\frac{\pi}{2} - \cos0 \right) + 2\left( \sin\frac{\pi}{2} - \sin0 \right)\]
\[ = - 4\left( 0 - 1 \right) + 2\left( 1 - 0 \right)\]
\[ = 4 + 2\]
\[ = 6\]