What is the integral of 6sin^2(x)cos^3(x) ?

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

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integral of 6sin^2(x)cos^3(x)

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

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

Solution steps

\int 6 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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 6 \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

= 6 (\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}

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

Substitute back

u = sin (x)

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

Add a constant to the solution

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

\int \frac{1}{- t ^{2} + 4 t + 5} d t

\int \frac{1}{5 - x} d x

\int 256 - x^{2} d x

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

\frac{d y}{d x} = \frac{3 x ^{2}}{e ^{2 y}}, y (0) = \frac{1}{2}

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