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

The integral of sin^5(θ)cos(θ) is (sin^6(θ))/6+C

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integral of sin^5(θ)cos(θ)

\int sin^{5} (θ) cos (θ) d θ

\frac{sin ^{6} ( θ )}{6} + C

Solution steps

\int sin^{5} (θ) cos (θ) d θ

Apply u-substitution

: \int u^{5} d u

= \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{u ^{6}}{6}

Substitute back

u = sin (θ)

= \frac{sin ^{6} ( θ )}{6}

Add a constant to the solution

= \frac{sin ^{6} ( θ )}{6} + C

x^{2} y^{2} = 16, (1, 4)

\frac{\partial}{\partial x} (5 - 3 x^{2} - y^{2})

\frac{d y}{d x} - x^{2} y = 0

\int x^{7} [0.7] d x

1 - x

What is the integral of sin^5(θ)cos(θ) ?
The integral of sin^5(θ)cos(θ) is (sin^6(θ))/6+C