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

The integral of 5t*sin^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 5t*sin^2(t)

\int 5 t \cdot 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

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

= 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

\frac{d}{d x} (arctan (e^{3 x}))

x \to 5 lim (\frac{2}{x - 5})

\frac{d}{d x} (tan (sin (x)))

\int \frac{1}{x ^{2} 3 x ^{2} + 4} d x

y = 1 n (x^{3})

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