how to integrate sin^6x

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Solution

We know that $\int \sin^{n} x d x = - \frac{1}{n} \sin^{n - 1} x \cos x + \frac{n - 1}{n} \int \sin^{n - 2} x d x$

So, $\int \sin^{6} x d x = - \frac{1}{6} \sin^{5} x \cos x + \frac{5}{6} \int \sin^{4} x d x$

$= - \frac{1}{6} \sin^{5} x \cos x + \frac{5}{6} (- \frac{1}{4} \sin^{3} x \cos x + \frac{3}{4} \int \sin^{2} x d x) = - \frac{1}{6} \sin^{5} x \cos x - \frac{5}{24} \sin^{3} x \cos x + \frac{5}{8} \int \sin^{2} x d x$

$= - \frac{1}{6} \sin^{5} x \cos x - \frac{5}{24} \sin^{3} x \cos x + \frac{5}{8} (- \frac{1}{2} \sin x \cos x + \frac{1}{2} \int \sin^{0} x d x) = - \frac{1}{6} \sin^{5} x \cos x - \frac{5}{24} \sin^{3} x \cos x + \frac{5}{8} (- \frac{1}{2} \sin x \cos x + \frac{1}{2} x) + C$

$= - \frac{1}{6} \sin^{5} x \cos x - \frac{5}{24} \sin^{3} x \cos x - \frac{5}{16} \sin x \cos x + \frac{5}{16} x + C = - \frac{1}{12} \sin 2 x \sin^{4} x - \frac{5}{48} \sin 2 x \sin^{2} x - \frac{5}{32} \sin 2 x + \frac{5}{16} x + C$

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