What is the integral of-6sin^2(t) ?

The integral of-6sin^2(t) is -3(t-1/2 sin(2t))+C

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integral of-6sin^2(t)

\int - 6 sin^{2} (t) d t

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

Solution steps

\int - 6 sin^{2} (t) d t

Take the constant out:

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

= - 6 \cdot \int sin^{2} (t) d t

Rewrite using trig identities

= - 6 \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

= - 6 \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

= - 6 \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)

= - 6 \cdot \frac{1}{2} (t - \frac{1}{2} sin (2 t))

Simplify

- 6 \cdot \frac{1}{2} (t - \frac{1}{2} sin (2 t)) : - 3 (t - \frac{1}{2} sin (2 t))

= - 3 (t - \frac{1}{2} sin (2 t))

Add a constant to the solution

= - 3 (t - \frac{1}{2} sin (2 t)) + C

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

x \to \infty lim (\frac{1}{1})

n = 0 \sum \infty \frac{n + 1}{2 n - 1}

f (x) = 5 x^{4} ln (x)

\int 1 - t^{- 0.52} d t

What is the integral of-6sin^2(t) ?
The integral of-6sin^2(t) is -3(t-1/2 sin(2t))+C