What is the integral of cos^5(4-x)sin(4-x) ?

The integral of cos^5(4-x)sin(4-x) is 1/6 cos^6(4-x)+C

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integral of cos^5(4-x)sin(4-x)

\int cos^{5} (4 - x) sin (4 - x) d x

\frac{1}{6} cos^{6} (4 - x) + C

Solution steps

\int cos^{5} (4 - x) sin (4 - x) d x

Simplify

cos^{5} (4 - x) : cos^{4} (4 - x) cos (4 - x)

= \int cos^{4} (4 - x) cos (4 - x) sin (4 - x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int - v^{5} d v

= - \int - v^{5} d v

Take the constant out:

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

= - (- \int v^{5} d v)

Apply the Power Rule

: \frac{v ^{6}}{6}

= - (- \frac{v ^{6}}{6})

Substitute back

= - (- \frac{cos ^{6} ( 4 - x )}{6})

Simplify

- (- \frac{cos ^{6} ( 4 - x )}{6}) : \frac{1}{6} cos^{6} (4 - x)

= \frac{1}{6} cos^{6} (4 - x)

Add a constant to the solution

= \frac{1}{6} cos^{6} (4 - x) + C

\int \frac{x ^{2} - 3 x + x}{x ^{2}} d x

\frac{d}{d x} (- \frac{6}{( x - 1 ) ^{4}})

\int e^{x - 2 x} sin (x) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} + e^{- 2 x}

\int_{0}^{4} \frac{1}{( x - 4 ) ^{2}} d x

What is the integral of cos^5(4-x)sin(4-x) ?
The integral of cos^5(4-x)sin(4-x) is 1/6 cos^6(4-x)+C