What is the integral of sin(2x-pi/4) ?

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

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integral of sin(2x-pi/4)

\int sin (2 x - \frac{π}{4}) d x

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

Solution steps

\int sin (2 x - \frac{π}{4}) 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

Use the common integral:

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

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

Substitute back

u = 2 x - \frac{π}{4}

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

Simplify

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

Add a constant to the solution

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

x \to 1 lim (1)

y^{^{'}} = \frac{x ^{2} + 1}{3 y ^{2}}

\int \frac{1}{x ^{2} - 7} d x

- 4 y^{^{''}} + 16 y = 0, y (1) = 2, y^{^{'}} (2) = 1

y = x^{3} (x - 1)^{2}

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