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

The integral of (2-sin(t/4))^2cos(t/4) is 16sin(t/4)+8cos^2(t/4)+4/3 sin^3(t/4)+C

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

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

16 sin (\frac{t}{4}) + 8 cos^{2} (\frac{t}{4}) + \frac{4}{3} sin^{3} (\frac{t}{4}) + C

Solution steps

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

Expand

(2 - sin (\frac{t}{4}))^{2} cos (\frac{t}{4}) : 4 cos (\frac{t}{4}) - 4 cos (\frac{t}{4}) sin (\frac{t}{4}) + sin^{2} (\frac{t}{4}) cos (\frac{t}{4})

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 4 cos (\frac{t}{4}) d t = 16 sin (\frac{t}{4})

\int 4 cos (\frac{t}{4}) sin (\frac{t}{4}) d t = - 8 cos^{2} (\frac{t}{4})

\int sin^{2} (\frac{t}{4}) cos (\frac{t}{4}) d t = \frac{4}{3} sin^{3} (\frac{t}{4})

= 16 sin (\frac{t}{4}) - (- 8 cos^{2} (\frac{t}{4})) + \frac{4}{3} sin^{3} (\frac{t}{4})

Simplify

= 16 sin (\frac{t}{4}) + 8 cos^{2} (\frac{t}{4}) + \frac{4}{3} sin^{3} (\frac{t}{4})

Add a constant to the solution

= 16 sin (\frac{t}{4}) + 8 cos^{2} (\frac{t}{4}) + \frac{4}{3} sin^{3} (\frac{t}{4}) + C

\frac{d y}{d x} = (cos (2 x))

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

\int (2 y sin (x) + e^{2 z}) d y

y^{^{''}} + 12 y^{^{'}} + 45 y = 0

\int_{1}^{2} \frac{9 x ^{3} + 4 x ^{2} - 2}{x} d x

What is the integral of (2-sin(t/4))^2cos(t/4) ?
The integral of (2-sin(t/4))^2cos(t/4) is 16sin(t/4)+8cos^2(t/4)+4/3 sin^3(t/4)+C