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

The integral of cos(x)sin^4(sin(x)) is -1/4 sin^3(sin(x))cos(sin(x))+3/8 (sin(x)-1/2 sin(2sin(x)))+C

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

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

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

Solution steps

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

Apply u-substitution

: \int sin^{4} (u) d u

= \int sin^{4} (u) d u

Apply Integral Reduction

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u

\int sin^{2} (u) d u = \frac{1}{2} (u - \frac{1}{2} sin (2 u))

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = sin (x)

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

Simplify

- \frac{cos ( sin ( x ) ) sin ^{3} ( sin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (sin (x) - \frac{1}{2} sin (2 sin (x))) : - \frac{1}{4} sin^{3} (sin (x)) cos (sin (x)) + \frac{3}{8} (sin (x) - \frac{1}{2} sin (2 sin (x)))

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

Add a constant to the solution

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

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What is the integral of cos(x)sin^4(sin(x)) ?
The integral of cos(x)sin^4(sin(x)) is -1/4 sin^3(sin(x))cos(sin(x))+3/8 (sin(x)-1/2 sin(2sin(x)))+C