What is the integral of cos^7(9-x)sin(9-x) ?

The integral of cos^7(9-x)sin(9-x) is 1/8 cos^8(9-x)+C

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integral of cos^7(9-x)sin(9-x)

\int cos^{7} (9 - x) sin (9 - x) d x

\frac{1}{8} cos^{8} (9 - x) + C

Solution steps

\int cos^{7} (9 - x) sin (9 - 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} ( 9 - x )}{8})

Simplify

- (- \frac{cos ^{8} ( 9 - x )}{8}) : \frac{1}{8} cos^{8} (9 - x)

= \frac{1}{8} cos^{8} (9 - x)

Add a constant to the solution

= \frac{1}{8} cos^{8} (9 - x) + C

v^{'} - 2 = 2 v^{2}

\frac{d}{d x} ((2 x + 7)^{3})

\frac{4}{s} (\frac{1}{6 + 5 s})

\int_{0}^{\frac{π}{2}} - 8 cos^{5} (x) d x

\frac{\partial}{\partial x} (3 x^{2} (5 x + 7 y))

What is the integral of cos^7(9-x)sin(9-x) ?
The integral of cos^7(9-x)sin(9-x) is 1/8 cos^8(9-x)+C