What is the integral of 8cos(θ)cos^5(sin(θ)) ?

The integral of 8cos(θ)cos^5(sin(θ)) is 8(sin(sin(θ))-(2sin^3(sin(θ)))/3+(sin^5(sin(θ)))/5)+C

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integral of 8cos(θ)cos^5(sin(θ))

\int 8 cos (θ) cos^{5} (sin (θ)) d θ

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

Solution steps

\int 8 cos (θ) cos^{5} (sin (θ)) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int cos (θ) cos^{5} (sin (θ)) d θ

Apply u-substitution

: \int cos^{5} (u) d u

= 8 \cdot \int cos^{5} (u) d u

Rewrite using trig identities

= 8 \cdot \int (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

= 8 \cdot \int (1 - v^{2})^{2} d v

Expand

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 8 \cdot \int 1 - 2 v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 8 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 8 (sin (sin (θ)) - \frac{2 sin ^{3} ( sin ( θ ) )}{3} + \frac{sin ^{5} ( sin ( θ ) )}{5})

Add a constant to the solution

= 8 (sin (sin (θ)) - \frac{2 sin ^{3} ( sin ( θ ) )}{3} + \frac{sin ^{5} ( sin ( θ ) )}{5}) + C

\frac{d}{d θ} (16 sin (θ))

\frac{d}{d x} (\frac{1}{5 x + 2})

(ln (3 x))^{2}

\int_{- 1}^{3} x^{3} + 2 x - 5 d x

\frac{\partial}{\partial x} (1 x + 14 y)

What is the integral of 8cos(θ)cos^5(sin(θ)) ?
The integral of 8cos(θ)cos^5(sin(θ)) is 8(sin(sin(θ))-(2sin^3(sin(θ)))/3+(sin^5(sin(θ)))/5)+C