What is the integral of sin^3(1/5 x)cos^2(1/5 x) ?

The integral of sin^3(1/5 x)cos^2(1/5 x) is -5/3 cos^3(x/5)+cos^5(x/5)+C

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integral of sin^3(1/5 x)cos^2(1/5 x)

\int sin^{3} (\frac{1}{5} x) cos^{2} (\frac{1}{5} x) d x

- \frac{5}{3} cos^{3} (\frac{x}{5}) + cos^{5} (\frac{x}{5}) + C

Solution steps

\int sin^{3} (\frac{1}{5} x) cos^{2} (\frac{1}{5} x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{2} (u) \cdot 5 d u

= \int sin^{3} (u) cos^{2} (u) \cdot 5 d u

Take the constant out:

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

= 5 \cdot \int sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

= 5 \cdot \int (1 - cos^{2} (u)) sin (u) cos^{2} (u) d u

Apply u-substitution

: \int - v^{2} (1 - v^{2}) d v

= 5 \cdot \int - v^{2} (1 - v^{2}) d v

Expand

- v^{2} (1 - v^{2}) : - v^{2} + v^{4}

= 5 \cdot \int - v^{2} + v^{4} 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

= 5 (- \int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 5 (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 5 (- \frac{cos ^{3} ( \frac{1}{5} x )}{3} + \frac{cos ^{5} ( \frac{1}{5} x )}{5})

Simplify

5 (- \frac{cos ^{3} ( \frac{1}{5} x )}{3} + \frac{cos ^{5} ( \frac{1}{5} x )}{5}) : - \frac{5}{3} cos^{3} (\frac{x}{5}) + cos^{5} (\frac{x}{5})

= - \frac{5}{3} cos^{3} (\frac{x}{5}) + cos^{5} (\frac{x}{5})

Add a constant to the solution

= - \frac{5}{3} cos^{3} (\frac{x}{5}) + cos^{5} (\frac{x}{5}) + C

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What is the integral of sin^3(1/5 x)cos^2(1/5 x) ?
The integral of sin^3(1/5 x)cos^2(1/5 x) is -5/3 cos^3(x/5)+cos^5(x/5)+C