What is the integral of sin^2(r) ?

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

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integral of sin^2(r)

\int sin^{2} (r) d r

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

Solution steps

\int sin^{2} (r) d r

Rewrite using trig identities

= \int \frac{1 - cos ( 2 r )}{2} d r

Take the constant out:

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

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

Apply the Sum Rule:

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

= \frac{1}{2} (\int 1 d r - \int cos (2 r) d r)

\int 1 d r = r

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

= \frac{1}{2} (r - \frac{1}{2} sin (2 r))

Add a constant to the solution

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

\int (6 csc (x) cot (x)) d x

x^{2} \cdot sin (x)

f (x) = 3 x^{2} - 2 x

- 9 x^{4}

\int sin (\frac{π x}{2}) d x

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