What is the integral of 2tsin^2(t) ?

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

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integral of 2tsin^2(t)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \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

= 2 (\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))

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

Simplify

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

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

Add a constant to the solution

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

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