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

The integral of sin(sqrt(2x-1)) is -cos(sqrt(2x-1))sqrt(2x-1)+sin(sqrt(2x-1))+C

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integral of sin(sqrt(2x-1))

\int sin (2 x - 1) d x

- cos (2 x - 1) 2 x - 1 + sin (2 x - 1) + C

Solution steps

\int sin (2 x - 1) d x

Apply u-substitution

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

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

Apply u-substitution

: \int sin (v) \cdot 2 v d v

= \frac{1}{2} \cdot \int sin (v) \cdot 2 v d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int sin (v) v d v

Apply Integration By Parts

: - v cos (v) - \int - cos (v) d v

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

\int - cos (v) d v = - sin (v)

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

Substitute back

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

Simplify

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

= - cos (2 x - 1) 2 x - 1 + sin (2 x - 1)

Add a constant to the solution

= - cos (2 x - 1) 2 x - 1 + sin (2 x - 1) + C

\int 3 (3 x + 12)^{5} d x

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\int \frac{1}{2} cos (2 t) d t

\int (ln (x) + sin (x) + 8) d x

What is the integral of sin(sqrt(2x-1)) ?
The integral of sin(sqrt(2x-1)) is -cos(sqrt(2x-1))sqrt(2x-1)+sin(sqrt(2x-1))+C