What is the integral of 16sin^2(x)cos^3(x) ?

The integral of 16sin^2(x)cos^3(x) is 16((sin^3(x))/3-(sin^5(x))/5)+C

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integral of 16sin^2(x)cos^3(x)

\int 16 sin^{2} (x) cos^{3} (x) d x

16 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

Solution steps

\int 16 sin^{2} (x) cos^{3} (x) d x

Take the constant out:

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

= 16 \cdot \int sin^{2} (x) cos^{3} (x) d x

Rewrite using trig identities

= 16 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{2} (x) d x

Apply u-substitution

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

= 16 \cdot \int u^{2} (1 - u^{2}) d u

Expand

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

= 16 \cdot \int u^{2} - u^{4} 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

= 16 (\int u^{2} d u - \int u^{4} d u)

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

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

= 16 (\frac{u ^{3}}{3} - \frac{u ^{5}}{5})

Substitute back

u = sin (x)

= 16 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5})

Add a constant to the solution

= 16 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

\frac{1}{4 x + 3}

\frac{\partial}{\partial x} (7 x - 3 y + 4)

\int_{0}^{1} x^{6} - x d x

y^{^{'}} + y \cdot cos (x) = sin (x) \cdot cos (x)

What is the integral of 16sin^2(x)cos^3(x) ?
The integral of 16sin^2(x)cos^3(x) is 16((sin^3(x))/3-(sin^5(x))/5)+C