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

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

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integral of+cos(θ)cos^5(sin(θ))

\int + cos (θ) cos^{5} (sin (θ)) d θ

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

Solution steps

\int cos (θ) cos^{5} (sin (θ)) d θ

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}

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

Substitute back

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

Add a constant to the solution

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

x^{^{'}} = \frac{1}{x ^{2}}, x (1) = 1

y = \frac{1}{a ^{2} - x ^{2}}

\frac{\partial}{\partial x} (e^{\frac{x}{b ( x )}})

t \to 0 lim (\frac{5}{1 + t})

y^{^{''}} - 3 y^{^{'}} = 18 t + 26 sin (2 t)

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