What is the integral of cos^5(9-x)sin(9-x) ?

The integral of cos^5(9-x)sin(9-x) is 1/6 cos^6(9-x)+C

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integral of cos^5(9-x)sin(9-x)

\int cos^{5} (9 - x) sin (9 - x) d x

\frac{1}{6} cos^{6} (9 - x) + C

Solution steps

\int cos^{5} (9 - x) sin (9 - x) d x

Simplify

cos^{5} (9 - x) : cos^{4} (9 - x) cos (9 - x)

= \int cos^{4} (9 - x) cos (9 - x) sin (9 - 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} ( 9 - x )}{6})

Simplify

- (- \frac{cos ^{6} ( 9 - x )}{6}) : \frac{1}{6} cos^{6} (9 - x)

= \frac{1}{6} cos^{6} (9 - x)

Add a constant to the solution

= \frac{1}{6} cos^{6} (9 - x) + C

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

\int 3 cot^{5} (θ) sin^{4} (θ) d θ

x^{2} - 2 x + 1

arctan (x - \frac{π}{2})

What is the integral of cos^5(9-x)sin(9-x) ?
The integral of cos^5(9-x)sin(9-x) is 1/6 cos^6(9-x)+C