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

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

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integral of 5tsin^2(t)

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

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

Take the constant out:

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

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

Apply Integration By Parts

= 5 (\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))

= 5 (\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

= 5 (\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

\int 12 x^{2} + 6 x - 5 d x

\int_{0}^{\frac{π}{4}} tan^{8} (θ) sec^{2} (θ) d θ

\frac{e ^{x}}{ln ( x )}

\frac{d}{d x} (x t)

x \to - 2 - lim (\frac{1}{x - 2})

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