What is the integral of 9tsin^2(t) ?

The integral of 9tsin^2(t) is 9(1/2 t(t-1/2 sin(2t))-1/2 ((t^2)/2+1/4 cos(2t)))+C

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integral of 9tsin^2(t)

\int 9 t sin^{2} (t) d t

9 (\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 9 t sin^{2} (t) d t

Take the constant out:

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

= 9 \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

= 9 (\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))

= 9 (\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

= 9 (\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

y^{^{'}} = \frac{6 x sec ( \frac{y}{x} ) + y}{x}

\frac{d}{d x} (- \frac{8 x}{x ^{2} + 1})

\frac{d}{d x} (ln (4) lo g_{4} (x) + π^{3})

x \to 2 lim (3.9 + 9.6)

n = 0 \sum \infty \frac{10}{2 ^{n}}

What is the integral of 9tsin^2(t) ?
The integral of 9tsin^2(t) is 9(1/2 t(t-1/2 sin(2t))-1/2 ((t^2)/2+1/4 cos(2t)))+C