What is the integral of sin^3(x/(10))cos^2(x/(10)) ?

The integral of sin^3(x/(10))cos^2(x/(10)) is 10(-(cos^3(x/(10)))/3+(cos^5(x/(10)))/5)+C

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integral of sin^3(x/(10))cos^2(x/(10))

\int sin^{3} (\frac{x}{10}) cos^{2} (\frac{x}{10}) d x

10 (- \frac{cos ^{3} ( \frac{x}{10} )}{3} + \frac{cos ^{5} ( \frac{x}{10} )}{5}) + C

Solution steps

\int sin^{3} (\frac{x}{10}) cos^{2} (\frac{x}{10}) d x

Apply u-substitution

: \int sin^{3} (u) cos^{2} (u) \cdot 10 d u

= \int sin^{3} (u) cos^{2} (u) \cdot 10 d u

Take the constant out:

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

= 10 \cdot \int sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

= 10 \cdot \int (1 - cos^{2} (u)) sin (u) cos^{2} (u) d u

Apply u-substitution

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

= 10 \cdot \int - v^{2} (1 - v^{2}) d v

Expand

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

= 10 \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

= 10 (- \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}

= 10 (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 10 (- \frac{cos ^{3} ( \frac{x}{10} )}{3} + \frac{cos ^{5} ( \frac{x}{10} )}{5})

Add a constant to the solution

= 10 (- \frac{cos ^{3} ( \frac{x}{10} )}{3} + \frac{cos ^{5} ( \frac{x}{10} )}{5}) + C

y^{^{''}} + 12 y^{^{'}} + 35 y = 70 t^{2} + 48 t + 4 + 48 e^{t}

h (t) = - 5 t^{2}

\frac{d}{d x} (\frac{5 sin ( x )}{3 + cos ( x )})

\int_{0}^{\frac{5}{2}} \frac{1}{25 + 4 x ^{2}} d x

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What is the integral of sin^3(x/(10))cos^2(x/(10)) ?
The integral of sin^3(x/(10))cos^2(x/(10)) is 10(-(cos^3(x/(10)))/3+(cos^5(x/(10)))/5)+C