What is the integral of sin^2(5r) ?

The integral of sin^2(5r) is 1/2 (r-1/10 sin(10r))+C

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integral of sin^2(5r)

\int sin^{2} (5 r) d r

\frac{1}{2} (r - \frac{1}{10} sin (10 r)) + C

Solution steps

\int sin^{2} (5 r) d r

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (10 r)) 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 (10 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 (10 r) d r)

\int 1 d r = r

\int cos (10 r) d r = \frac{1}{10} sin (10 r)

= \frac{1}{2} (r - \frac{1}{10} sin (10 r))

Add a constant to the solution

= \frac{1}{2} (r - \frac{1}{10} sin (10 r)) + C

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What is the integral of sin^2(5r) ?
The integral of sin^2(5r) is 1/2 (r-1/10 sin(10r))+C