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

The integral of sin^4(t)cos^4(t) is

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integral of sin^4(t)cos^4(t)

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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} (t) d t - \int 2 sin^{6} (t) d t + \int sin^{8} (t) d t

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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