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

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

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integral of sin^2(θ)cos^5(θ)

\int sin^{2} (θ) cos^{5} (θ) d θ

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

Solution steps

\int sin^{2} (θ) cos^{5} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}

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

Substitute back

u = sin (θ)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{5 x y})

\frac{d}{d x} (sec^{2} (x) tan^{2} (x))

\int \frac{2 x + 4}{x ^{2} - x} d x

\frac{\partial}{\partial x} (\frac{1}{3} x^{\frac{3}{2}})

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