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

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

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

integral

sin^{2} (x) cos^{3} (x)

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= \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

= \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}

= \frac{u ^{3}}{3} - \frac{u ^{5}}{5}

Substitute back

u = sin (x)

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

Add a constant to the solution

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

in t e g r a l \frac{1}{v ^{4}}

in t e g r a l \frac{( x ^{3} + 2 )}{( 4 - x ^{2} )}

in t e g r a l x^{2} (x - 4)^{\frac{1}{2}}

in t e g r a l x^{2} \cdot (e^{x})^{2}

in t e g r a l e^{x} - 2^{x}

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