Question

\[\int\limits_{- \pi/2}^{\pi/2} \sin^4 x\ dx\]

Answer 1

\[Let\ I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^4 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \sin^2 x \right)^2 dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \frac{1 - \cos2x}{2} \right)^2 dx\]
\[ = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 - 2\cos2x + \cos^2 2x \right) dx\]
\[ = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} dx - \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos2x dx + \frac{1}{8} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 + \cos4x \right) dx\]
\[ = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} dx - \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos2x dx + \frac{1}{8} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} dx + \frac{1}{8} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos4x\ dx\]
\[ = \frac{3}{8} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} dx - \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos2x dx + \frac{1}{8} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos4x\ dx\]
\[ = \frac{3}{8} \left[ x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} - \frac{1}{4} \left[ \sin2x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} + \frac{1}{32} \left[ \sin4x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = \frac{3}{8}\left( \frac{\pi}{2} + \frac{\pi}{2} \right) - \frac{1}{4}\left( 0 - 0 \right) + \frac{1}{32}\left( 0 - 0 \right)\]
\[Hence\ I = \frac{3\pi}{8}\]