Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
✓
Answer
Let $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
Here $\text{f(x)}=\sin^2\text{x}$
$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function
Therefore,
$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
Here $\text{f(x)}=\sin^2\text{x}$
$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function
Therefore,
$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
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