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

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

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integral of t*sin^2(t)

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

\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 t sin^{2} (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

= \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))

= \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

= \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 t*sin^2(t) ?
The integral of t*sin^2(t) is 1/2 t(t-1/2 sin(2t))-1/2 ((t^2)/2+1/4 cos(2t))+C