What is the integral of 6sin^2(x)cos^5(x) ?

The integral of 6sin^2(x)cos^5(x) is 6((sin^3(x))/3-(2sin^5(x))/5+(sin^7(x))/7)+C

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integral of 6sin^2(x)cos^5(x)

\int 6 sin^{2} (x) cos^{5} (x) d x

6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}) + C

Solution steps

\int 6 sin^{2} (x) cos^{5} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int sin^{2} (x) cos^{5} (x) d x

Rewrite using trig identities

= 6 \cdot \int (1 - sin^{2} (x))^{2} cos (x) sin^{2} (x) d x

Apply u-substitution

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

= 6 \cdot \int u^{2} (1 - u^{2})^{2} d u

Expand

u^{2} (1 - u^{2})^{2} : u^{2} - 2 u^{4} + u^{6}

= 6 \cdot \int u^{2} - 2 u^{4} + u^{6} 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

= 6 (\int u^{2} d u - \int 2 u^{4} d u + \int u^{6} d u)

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= 6 (\frac{u ^{3}}{3} - \frac{2 u ^{5}}{5} + \frac{u ^{7}}{7})

Substitute back

u = sin (x)

= 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7})

Add a constant to the solution

= 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}) + C

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What is the integral of 6sin^2(x)cos^5(x) ?
The integral of 6sin^2(x)cos^5(x) is 6((sin^3(x))/3-(2sin^5(x))/5+(sin^7(x))/7)+C