What is the integral of (4-sin(t/5))^2cos(t/5) ?

The integral of (4-sin(t/5))^2cos(t/5) is 80sin(t/5)+20cos^2(t/5)+5/3 sin^3(t/5)+C

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integral of (4-sin(t/5))^2cos(t/5)

\int (4 - sin (\frac{t}{5}))^{2} cos (\frac{t}{5}) d t

80 sin (\frac{t}{5}) + 20 cos^{2} (\frac{t}{5}) + \frac{5}{3} sin^{3} (\frac{t}{5}) + C

Solution steps

\int (4 - sin (\frac{t}{5}))^{2} cos (\frac{t}{5}) d t

4 - sin (\frac{t}{5}) = 4 (1 - \frac{sin ( \frac{t}{5} )}{4})

= \int (4 (1 - \frac{sin ( \frac{t}{5} )}{4}))^{2} cos (\frac{t}{5}) d t

Simplify

(4 (1 - \frac{sin ( \frac{t}{5} )}{4}))^{2} cos (\frac{t}{5}) : 16 cos (\frac{t}{5}) - 8 sin (\frac{t}{5}) cos (\frac{t}{5}) + sin^{2} (\frac{t}{5}) cos (\frac{t}{5})

= \int 16 cos (\frac{t}{5}) - 8 sin (\frac{t}{5}) cos (\frac{t}{5}) + sin^{2} (\frac{t}{5}) cos (\frac{t}{5}) 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 16 cos (\frac{t}{5}) d t - \int 8 sin (\frac{t}{5}) cos (\frac{t}{5}) d t + \int sin^{2} (\frac{t}{5}) cos (\frac{t}{5}) d t

\int 16 cos (\frac{t}{5}) d t = 80 sin (\frac{t}{5})

\int 8 sin (\frac{t}{5}) cos (\frac{t}{5}) d t = - 20 cos^{2} (\frac{t}{5})

\int sin^{2} (\frac{t}{5}) cos (\frac{t}{5}) d t = \frac{5}{3} sin^{3} (\frac{t}{5})

= 80 sin (\frac{t}{5}) - (- 20 cos^{2} (\frac{t}{5})) + \frac{5}{3} sin^{3} (\frac{t}{5})

Simplify

= 80 sin (\frac{t}{5}) + 20 cos^{2} (\frac{t}{5}) + \frac{5}{3} sin^{3} (\frac{t}{5})

Add a constant to the solution

= 80 sin (\frac{t}{5}) + 20 cos^{2} (\frac{t}{5}) + \frac{5}{3} sin^{3} (\frac{t}{5}) + C

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What is the integral of (4-sin(t/5))^2cos(t/5) ?
The integral of (4-sin(t/5))^2cos(t/5) is 80sin(t/5)+20cos^2(t/5)+5/3 sin^3(t/5)+C