What is the integral of (sin(x)+cos^2(x))^3 ?

The integral of (sin(x)+cos^2(x))^3 is

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integral of (sin(x)+cos^2(x))^3

\int (sin (x) + cos^{2} (x))^{3} d x

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

Solution steps

\int (sin (x) + cos^{2} (x))^{3} d x

Expand

(sin (x) + cos^{2} (x))^{3} : sin^{3} (x) + 3 sin^{2} (x) cos^{2} (x) + 3 cos^{4} (x) sin (x) + cos^{6} (x)

= \int sin^{3} (x) + 3 sin^{2} (x) cos^{2} (x) + 3 cos^{4} (x) sin (x) + cos^{6} (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^{3} (x) d x + \int 3 sin^{2} (x) cos^{2} (x) d x + \int 3 cos^{4} (x) sin (x) d x + \int cos^{6} (x) d x

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

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

\int 3 cos^{4} (x) sin (x) d x = - \frac{3}{5} cos^{5} (x)

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

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

Add a constant to the solution

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

\int \frac{x ^{3} - 3 x - 8}{x - 4} d x

\int \frac{1}{x ^{4} + 16 x ^{2}} d x

\int \frac{x ^{2}}{2 x - x ^{2}} d x

\int \frac{x + 3}{6 x ^{3} - 7 x ^{2} - 3 x} d x

\int \frac{1}{a x - x ^{2}} d x

What is the integral of (sin(x)+cos^2(x))^3 ?
The integral of (sin(x)+cos^2(x))^3 is