What is the integral of-3sin^2(x) ?

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

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integral of-3sin^2(x)

\int - 3 sin^{2} (x) d x

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

Solution steps

\int - 3 sin^{2} (x) d x

Take the constant out:

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

= - 3 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

= - 3 \cdot \int \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

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

= - 3 \cdot \frac{1}{2} \cdot \int 1 - cos (2 x) d x

Apply the Sum Rule:

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

= - 3 \cdot \frac{1}{2} (\int 1 d x - \int cos (2 x) d x)

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

y = x^{- 5 c o s (x)}

x ln (x) - x

\int \frac{x + 4}{x ^{2} + 6 x - 16} d x

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

y = cos (x), at x = - \frac{π}{2}

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