What is the integral of cos^5(x)sin^1(x) ?

The integral of cos^5(x)sin^1(x) is (sin^2(x))/2-(sin^4(x))/2+(sin^6(x))/6+C

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integral of cos^5(x)sin^1(x)

\int cos^{5} (x) sin^{1} (x) d x

\frac{sin ^{2} ( x )}{2} - \frac{sin ^{4} ( x )}{2} + \frac{sin ^{6} ( x )}{6} + C

Solution steps

\int cos^{5} (x) sin^{1} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

Expand

u (1 - u^{2})^{2} : u - 2 u^{3} + u^{5}

= \int u - 2 u^{3} + u^{5} d u

Apply the Sum Rule:

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

= \int u d u - \int 2 u^{3} d u + \int u^{5} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 u^{3} d u = \frac{u ^{4}}{2}

\int u^{5} d u = \frac{u ^{6}}{6}

= \frac{u ^{2}}{2} - \frac{u ^{4}}{2} + \frac{u ^{6}}{6}

Substitute back

u = sin (x)

= \frac{sin ^{2} ( x )}{2} - \frac{sin ^{4} ( x )}{2} + \frac{sin ^{6} ( x )}{6}

Add a constant to the solution

= \frac{sin ^{2} ( x )}{2} - \frac{sin ^{4} ( x )}{2} + \frac{sin ^{6} ( x )}{6} + C

What is the integral of cos^5(x)sin^1(x) ?
The integral of cos^5(x)sin^1(x) is (sin^2(x))/2-(sin^4(x))/2+(sin^6(x))/6+C