What is the integral of sin^2(8x) ?

The integral of sin^2(8x) is 1/2 (x-1/16 sin(16x))+C

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integral of sin^2(8x)

\int sin^{2} (8 x) d x

\frac{1}{2} (x - \frac{1}{16} sin (16 x)) + C

Solution steps

\int sin^{2} (8 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (16 x)) d x

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 (16 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

= \frac{1}{2} (\int 1 d x - \int cos (16 x) d x)

\int 1 d x = x

\int cos (16 x) d x = \frac{1}{16} sin (16 x)

= \frac{1}{2} (x - \frac{1}{16} sin (16 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{16} sin (16 x)) + C

\frac{d}{d x} (5 x^{4} - 60 x^{2} + 64)

\int e^{x} x^{- 1} d x

\int_{2}^{\infty} e^{- 8 x} d x

y^{^{'}} = e^{x} y

\frac{d}{d x} (3 x - x^{3})

What is the integral of sin^2(8x) ?
The integral of sin^2(8x) is 1/2 (x-1/16 sin(16x))+C