What is the integral of (4-sin(t/2))^2cos(t/2) ?

The integral of (4-sin(t/2))^2cos(t/2) is -2/3 (4-sin(t/2))^3+C

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integral of (4-sin(t/2))^2cos(t/2)

\int (4 - sin (\frac{t}{2}))^{2} cos (\frac{t}{2}) d t

- \frac{2}{3} (4 - sin (\frac{t}{2}))^{3} + C

Solution steps

\int (4 - sin (\frac{t}{2}))^{2} cos (\frac{t}{2}) d t

Apply u-substitution

: \int - 2 u^{2} d u

= \int - 2 u^{2} d u

Take the constant out:

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

= - 2 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = 4 - sin (\frac{t}{2})

= - 2 \cdot \frac{( 4 - sin ( \frac{t}{2} ) ) ^{3}}{3}

Simplify

- 2 \cdot \frac{( 4 - sin ( \frac{t}{2} ) ) ^{3}}{3} : - \frac{2}{3} (4 - sin (\frac{t}{2}))^{3}

= - \frac{2}{3} (4 - sin (\frac{t}{2}))^{3}

Add a constant to the solution

= - \frac{2}{3} (4 - sin (\frac{t}{2}))^{3} + C

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What is the integral of (4-sin(t/2))^2cos(t/2) ?
The integral of (4-sin(t/2))^2cos(t/2) is -2/3 (4-sin(t/2))^3+C