What is the integral of sin^2(φ)cos^6(φ) ?

The integral of sin^2(φ)cos^6(φ) is

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integral of sin^2(φ)cos^6(φ)

\int sin^{2} (φ) cos^{6} (φ) d φ

\frac{5}{192} cos^{3} (φ) sin (φ) + \frac{1}{6} cos^{5} (φ) sin (φ) + \frac{5}{32} (2 φ + sin (2 φ)) - \frac{1}{8} cos^{7} (φ) sin (φ) - \frac{7}{48} cos^{5} (φ) sin (φ) - \frac{35}{256} (2 φ + sin (2 φ)) + C

Solution steps

\int sin^{2} (φ) cos^{6} (φ) d φ

Rewrite using trig identities

= \int (1 - cos^{2} (φ)) cos^{6} (φ) d φ

Expand

(1 - cos^{2} (φ)) cos^{6} (φ) : cos^{6} (φ) - cos^{8} (φ)

= \int cos^{6} (φ) - cos^{8} (φ) d φ

Apply the Sum Rule:

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

= \int cos^{6} (φ) d φ - \int cos^{8} (φ) d φ

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

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

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

Simplify

\frac{sin ( φ ) cos ^{5} ( φ )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (φ) sin (φ) + \frac{3}{8} (φ + \frac{1}{2} sin (2 φ))) - (\frac{sin ( φ ) cos ^{7} ( φ )}{8} + \frac{7}{8} (\frac{sin ( φ ) cos ^{5} ( φ )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (φ) sin (φ) + \frac{3}{8} (φ + \frac{1}{2} sin (2 φ))))) : \frac{5}{192} cos^{3} (φ) sin (φ) + \frac{1}{6} cos^{5} (φ) sin (φ) + \frac{5}{32} (2 φ + sin (2 φ)) - \frac{1}{8} cos^{7} (φ) sin (φ) - \frac{7}{48} cos^{5} (φ) sin (φ) - \frac{35}{256} (2 φ + sin (2 φ))

= \frac{5}{192} cos^{3} (φ) sin (φ) + \frac{1}{6} cos^{5} (φ) sin (φ) + \frac{5}{32} (2 φ + sin (2 φ)) - \frac{1}{8} cos^{7} (φ) sin (φ) - \frac{7}{48} cos^{5} (φ) sin (φ) - \frac{35}{256} (2 φ + sin (2 φ))

Add a constant to the solution

= \frac{5}{192} cos^{3} (φ) sin (φ) + \frac{1}{6} cos^{5} (φ) sin (φ) + \frac{5}{32} (2 φ + sin (2 φ)) - \frac{1}{8} cos^{7} (φ) sin (φ) - \frac{7}{48} cos^{5} (φ) sin (φ) - \frac{35}{256} (2 φ + sin (2 φ)) + C