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

The integral of sin^3(θ)cos^6(θ) is -(cos^7(θ))/7+(cos^9(θ))/9+C

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integral of sin^3(θ)cos^6(θ)

\int sin^{3} (θ) cos^{6} (θ) d θ

- \frac{cos ^{7} ( θ )}{7} + \frac{cos ^{9} ( θ )}{9} + C

Solution steps

\int sin^{3} (θ) cos^{6} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{6} (1 - u^{2}) : - u^{6} + u^{8}

= \int - u^{6} + u^{8} 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^{6} d u + \int u^{8} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= - \frac{u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = cos (θ)

= - \frac{cos ^{7} ( θ )}{7} + \frac{cos ^{9} ( θ )}{9}

Add a constant to the solution

= - \frac{cos ^{7} ( θ )}{7} + \frac{cos ^{9} ( θ )}{9} + C

f (2.1) = 3 x^{2} - 4 x + 1, (2)

\int ln ∣ x ∣ cos (x) d x

\frac{\partial}{\partial x} (5 \cdot (y + 4) \cdot (x - 2)^{3})

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

\frac{d}{d x} (\frac{4 x ^{2} + 3 x}{2 x - 7})

What is the integral of sin^3(θ)cos^6(θ) ?
The integral of sin^3(θ)cos^6(θ) is -(cos^7(θ))/7+(cos^9(θ))/9+C