What is the integral of 40sin^2(5x) ?

The integral of 40sin^2(5x) is 20(x-1/10 sin(10x))+C

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integral of 40sin^2(5x)

\int 40 sin^{2} (5 x) d x

20 (x - \frac{1}{10} sin (10 x)) + C

Solution steps

\int 40 sin^{2} (5 x) d x

Take the constant out:

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

= 40 \cdot \int sin^{2} (5 x) d x

Rewrite using trig identities

= 40 \cdot \int \frac{1}{2} (1 - cos (10 x)) d x

Take the constant out:

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

= 40 \cdot \frac{1}{2} \cdot \int 1 - cos (10 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

= 40 \cdot \frac{1}{2} (\int 1 d x - \int cos (10 x) d x)

\int 1 d x = x

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

= 40 \cdot \frac{1}{2} (x - \frac{1}{10} sin (10 x))

Simplify

40 \cdot \frac{1}{2} (x - \frac{1}{10} sin (10 x)) : 20 (x - \frac{1}{10} sin (10 x))

= 20 (x - \frac{1}{10} sin (10 x))

Add a constant to the solution

= 20 (x - \frac{1}{10} sin (10 x)) + C

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What is the integral of 40sin^2(5x) ?
The integral of 40sin^2(5x) is 20(x-1/10 sin(10x))+C