What is the integral of 2cos(x)cos^5(sin(x)) ?

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

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integral of 2cos(x)cos^5(sin(x))

\int 2 cos (x) cos^{5} (sin (x)) d x

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

Solution steps

\int 2 cos (x) cos^{5} (sin (x)) d x

Take the constant out:

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

= 2 \cdot \int cos (x) cos^{5} (sin (x)) d x

Apply u-substitution

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

= 2 \cdot \int cos^{5} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 2 \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

= 2 (\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}

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

Substitute back

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

Add a constant to the solution

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

\frac{d}{d x} (5 \frac{x y}{2})

2 x \frac{d y}{d x} = cos^{2} (y)

arccos (x^{\frac{1}{2}})

4 x^{3} + e^{16 x} - 3^{2 x - 1}

\int (\frac{x ^{3}}{1 - x ^{2}}) d x

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