Evaluate the following integrals: ^ 2 _ - 2 | |dx

^ 2 _ - 2 | |dx = ^ 0 _ - 4 | |dx+ _ 0 ^ 2 | |dx = ^ 0 _ - 4 (- )dx+ _ 0 ^ 2 ( )dx | |= ,&0 2 - ,&- 4 0 = ^ 0 _ - 4 ^2x dx+ _ 0 ^ 2 ^2x dx =- ^ 0 _ - 4 1- 2x 2 dx+ _ 0 ^ 2 1- 2x 2 dx =-1 2 ^ 0 _ - 4 dx+1 2 ^ 0 _ - 4 2x dx+1 2 _ 0 ^ 2 dx-1 2 _ 0 ^ 2 2x dx =-1 2 [x ]^0_ 4 +1 2 [ 2x 2 ]^0_ 2 +1 2 [x ]^ 2 _0-1 2 [ 2x 2 ]^ 2 _0 =-1 2 (0+ 4 )+1 4 (0+1)+ 4 -1 4 (0-0) = 8 +1 4

Evaluate the following integrals: ^ 2 _ - 2

Download App

Question

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$

✓

Answer

$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}|\sin\text{x}|\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}(-\sin\text{x})\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}(\sin\text{x})\text{dx}$ $\begin{pmatrix}|\sin\text{x}|=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\end{cases} \end{pmatrix} $
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin^2\text{x dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin^2\text{x dx}$
$=-\int\limits^{0}_{-\frac{\pi}{4}}\frac{1-\cos2\text{x}}{2}\text{ dx}+\int\limits_{0}^{\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{ dx}$
$=-\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\text{dx}+\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\cos2\text{x dx}+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x dx}$
$=-\frac{1}{2}\times\big[\text{x}\big]^0_\frac{\pi}{4}+\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^0_{\frac{\pi}{2}}+\frac{1}{2}\times\big[\text{x}\big]^{\frac{\pi}{2}}_0-\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=-\frac{1}{2}\Big(0+\frac{\pi}{4}\Big)+\frac{1}{4}(0+1)+\frac{\pi}{4}-\frac{1}{4}(0-0)$
$=\frac{\pi}{8}+\frac{1}{4}$