What is the integral of cos^7(8-x)sin(8-x) ?

The integral of cos^7(8-x)sin(8-x) is 1/8 cos^8(8-x)+C

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integral of cos^7(8-x)sin(8-x)

\int cos^{7} (8 - x) sin (8 - x) d x

\frac{1}{8} cos^{8} (8 - x) + C

Solution steps

\int cos^{7} (8 - x) sin (8 - x) d x

Apply u-substitution

: \int - cos^{7} (u) sin (u) d u

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

Apply u-substitution

: \int - v^{7} d v

= - \int - v^{7} d v

Take the constant out:

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

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

Apply the Power Rule

: \frac{v ^{8}}{8}

= - (- \frac{v ^{8}}{8})

Substitute back

= - (- \frac{cos ^{8} ( 8 - x )}{8})

Simplify

- (- \frac{cos ^{8} ( 8 - x )}{8}) : \frac{1}{8} cos^{8} (8 - x)

= \frac{1}{8} cos^{8} (8 - x)

Add a constant to the solution

= \frac{1}{8} cos^{8} (8 - x) + C

y^{^{'}} - 4 y = 9 e^{7 t}

\int tan^{2} (8 x) sec^{4} (8 x) d x

\int y^{2} 4 y^{2} + 1 d y

\int_{0}^{1} x - x d x

\frac{d x}{d t} = sin (2 π (t + 3)), x (3) = 1

What is the integral of cos^7(8-x)sin(8-x) ?
The integral of cos^7(8-x)sin(8-x) is 1/8 cos^8(8-x)+C