What is the integral of cos(x)(1-cos(x))^{99}sin(x) ?

The integral of cos(x)(1-cos(x))^{99}sin(x) is -1/101 (1-cos(x))^{101}+1/100 (1-cos(x))^{100}+C

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integral of cos(x)(1-cos(x))^{99}sin(x)

\int cos (x) (1 - cos (x))^{99} sin (x) d x

- \frac{1}{101} (1 - cos (x))^{101} + \frac{1}{100} (1 - cos (x))^{100} + C

Solution steps

\int cos (x) (1 - cos (x))^{99} sin (x) d x

Apply u-substitution

: \int - u (1 - u)^{99} d u

= \int - u (1 - u)^{99} d u

Take the constant out:

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

= - \int u (1 - u)^{99} d u

Apply u-substitution

: \int - (- v + 1) v^{99} d v

= - \int - (- v + 1) v^{99} d v

Take the constant out:

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

= - (- \int (- v + 1) v^{99} d v)

Expand

(- v + 1) v^{99} : - v^{100} + v^{99}

= - (- \int - v^{100} + v^{99} 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

= - (- (- \int v^{100} d v + \int v^{99} d v))

\int v^{100} d v = \frac{v ^{101}}{101}

\int v^{99} d v = \frac{v ^{100}}{100}

= - (- (- \frac{v ^{101}}{101} + \frac{v ^{100}}{100}))

Substitute back

= - (- (- \frac{( 1 - cos ( x ) ) ^{101}}{101} + \frac{( 1 - cos ( x ) ) ^{100}}{100}))

Simplify

- (- (- \frac{( 1 - cos ( x ) ) ^{101}}{101} + \frac{( 1 - cos ( x ) ) ^{100}}{100})) : - \frac{1}{101} (1 - cos (x))^{101} + \frac{1}{100} (1 - cos (x))^{100}

= - \frac{1}{101} (1 - cos (x))^{101} + \frac{1}{100} (1 - cos (x))^{100}

Add a constant to the solution

= - \frac{1}{101} (1 - cos (x))^{101} + \frac{1}{100} (1 - cos (x))^{100} + C

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What is the integral of cos(x)(1-cos(x))^{99}sin(x) ?
The integral of cos(x)(1-cos(x))^{99}sin(x) is -1/101 (1-cos(x))^{101}+1/100 (1-cos(x))^{100}+C