What is the integral of sin^4(x)*cos^4(x) ?

The integral of sin^4(x)*cos^4(x) is

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integral of sin^4(x)*cos^4(x)

\int sin^{4} (x) \cdot cos^{4} (x) d x

- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (x) cos (x) + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{1}{8} sin^{7} (x) cos (x) + C

Solution steps

\int sin^{4} (x) cos^{4} (x) d x

Rewrite using trig identities

= \int sin^{4} (x) (1 - sin^{2} (x))^{2} d x

Expand

sin^{4} (x) (1 - sin^{2} (x))^{2} : sin^{4} (x) - 2 sin^{6} (x) + sin^{8} (x)

= \int sin^{4} (x) - 2 sin^{6} (x) + sin^{8} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sin^{4} (x) d x - \int 2 sin^{6} (x) d x + \int sin^{8} (x) d x

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

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

\int sin^{8} (x) d x = - \frac{cos ( x ) sin ^{7} ( x )}{8} + \frac{7}{8} (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))))

= - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - 2 (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{cos ( x ) sin ^{7} ( x )}{8} + \frac{7}{8} (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))))

Simplify

- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - 2 (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{cos ( x ) sin ^{7} ( x )}{8} + \frac{7}{8} (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) : - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (x) cos (x) + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{1}{8} sin^{7} (x) cos (x)

= - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (x) cos (x) + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{1}{8} sin^{7} (x) cos (x)

Add a constant to the solution

= - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (x) cos (x) + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))) - \frac{1}{8} sin^{7} (x) cos (x) + C

What is the integral of sin^4(x)*cos^4(x) ?
The integral of sin^4(x)*cos^4(x) is