What is the integral of sin^2(1-2x) ?

The integral of sin^2(1-2x) is 1/2 (x+1/4 sin(2-4x))+C

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integral of sin^2(1-2x)

\int sin^{2} (1 - 2 x) d x

\frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x)) + C

Solution steps

\int sin^{2} (1 - 2 x) d x

Rewrite using trig identities

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

Expand

2 (1 - 2 x) : 2 - 4 x

= \frac{1}{2} \cdot \int 1 - cos (2 - 4 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 (2 - 4 x) d x)

\int 1 d x = x

\int cos (2 - 4 x) d x = - \frac{1}{4} sin (2 - 4 x)

= \frac{1}{2} (x - (- \frac{1}{4} sin (2 - 4 x)))

Simplify

= \frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x))

Add a constant to the solution

= \frac{1}{2} (x + \frac{1}{4} sin (2 - 4 x)) + C

What is the integral of sin^2(1-2x) ?
The integral of sin^2(1-2x) is 1/2 (x+1/4 sin(2-4x))+C