Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
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Answer
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{x}\big)^2\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big(\frac{1-\cos2\text{x}}{2}\Big)^2\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(1-2\cos2\text{x}+\cos^22\text{x}\big)\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{3}{8}\Big[\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{4}\Big[\sin2\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\frac{1}{32}\Big[\sin4\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{3}{8}\Big(\frac{\pi}{2}+\frac{\pi}{2}\Big)-\frac{1}{4}(0-0)+\frac{1}{32}(0-0)$
Hence, $\text{I}=\frac{3\pi}{8}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{x}\big)^2\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big(\frac{1-\cos2\text{x}}{2}\Big)^2\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(1-2\cos2\text{x}+\cos^22\text{x}\big)\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{3}{8}\Big[\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{4}\Big[\sin2\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\frac{1}{32}\Big[\sin4\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{3}{8}\Big(\frac{\pi}{2}+\frac{\pi}{2}\Big)-\frac{1}{4}(0-0)+\frac{1}{32}(0-0)$
Hence, $\text{I}=\frac{3\pi}{8}$
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