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

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

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integral of sin^2(x-1)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int (1 - cos (u)) \frac{1}{2} d u

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

Take the constant out:

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

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

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

\int 1 d u = u

\int cos (u) d u = sin (u)

= \frac{1}{2} \cdot \frac{1}{2} (u - sin (u))

Substitute back

u = 2 (x - 1)

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

Simplify

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

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

Add a constant to the solution

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

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