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

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

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integral of-9sin^2(t)

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

- \frac{9}{2} (t - \frac{1}{2} sin (2 t)) + C

Solution steps

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

Rewrite using trig identities

= - 9 \cdot \int \frac{1 - cos ( 2 t )}{2} d t

Take the constant out:

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

= - 9 \cdot \frac{1}{2} \cdot \int 1 - cos (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 9 \cdot \frac{1}{2} (\int 1 d t - \int cos (2 t) d t)

\int 1 d t = t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

= - 9 \cdot \frac{1}{2} (t - \frac{1}{2} sin (2 t))

Simplify

- 9 \cdot \frac{1}{2} (t - \frac{1}{2} sin (2 t)) : - \frac{9}{2} (t - \frac{1}{2} sin (2 t))

= - \frac{9}{2} (t - \frac{1}{2} sin (2 t))

Add a constant to the solution

= - \frac{9}{2} (t - \frac{1}{2} sin (2 t)) + C

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