What is the integral of 7tsin^2(t) ?

The integral of 7tsin^2(t) is 7(1/2 t(t-1/2 sin(2t))-1/2 ((t^2)/2+1/4 cos(2t)))+C

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integral of 7tsin^2(t)

\int 7 t sin^{2} (t) d t

7 (\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 7 t sin^{2} (t) d t

Take the constant out:

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

= 7 \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

= 7 (\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))

= 7 (\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

= 7 (\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{\partial}{\partial y} (arctan (\frac{x}{y}))

\frac{\partial}{\partial y} (5 arctan (\frac{x}{y}))

\frac{d y}{d x} = 8 y x

y^{^{'}} = 5 x 3 y + 11, y (0) = 3

f (x) = ln (ln (3 x))

What is the integral of 7tsin^2(t) ?
The integral of 7tsin^2(t) is 7(1/2 t(t-1/2 sin(2t))-1/2 ((t^2)/2+1/4 cos(2t)))+C