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

The integral of cos^2(t)sin^5(t is sin^5(d)(1/2 t(t+1/2 sin(2t))-1/2 ((t^2)/2-1/4 cos(2t)))+C

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integral of cos^2(t)sin^5(t

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{2} t (t + \frac{1}{2} sin (2 t)) - \int \frac{1}{2} (t + \frac{1}{2} sin (2 t)) d t

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

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

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

Add a constant to the solution

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

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What is the integral of cos^2(t)sin^5(t ?
The integral of cos^2(t)sin^5(t is sin^5(d)(1/2 t(t+1/2 sin(2t))-1/2 ((t^2)/2-1/4 cos(2t)))+C