What is the integral of 7cos(θ)cos^5(sin(θ)) ?

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

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integral of 7cos(θ)cos^5(sin(θ))

\int 7 cos (θ) cos^{5} (sin (θ)) d θ

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

Solution steps

\int 7 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

= 7 \cdot \int cos (θ) cos^{5} (sin (θ)) d θ

Apply u-substitution

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

= 7 \cdot \int cos^{5} (u) d u

Rewrite using trig identities

= 7 \cdot \int (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

= 7 \cdot \int (1 - v^{2})^{2} d v

Expand

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

= 7 \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

= 7 (\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}

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

Substitute back

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

Add a constant to the solution

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

arccos (7 x^{3})

\frac{\partial}{\partial x} (2 y e^{x^{2} + y^{2}})

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

f (x) = x^{2} + 11

t \to \frac{π}{3} lim (cos (t) + 2)

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