What is the integral of sin((1-x)/3) ?

The integral of sin((1-x)/3) is 3cos(1/3 (1-x))+C

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integral of sin((1-x)/3)

\int sin (\frac{1 - x}{3}) d x

3 cos (\frac{1}{3} (1 - x)) + C

Solution steps

\int sin (\frac{1 - x}{3}) d x

Apply u-substitution

: \int - sin (\frac{u}{3}) d u

= \int - sin (\frac{u}{3}) d u

Take the constant out:

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

= - \int sin (\frac{u}{3}) d u

Apply u-substitution

: \int sin (v) \cdot 3 d v

= - \int sin (v) \cdot 3 d v

Take the constant out:

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

= - 3 \cdot \int sin (v) d v

Use the common integral:

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

= - 3 (- cos (v))

Substitute back

= - 3 (- cos (\frac{1 - x}{3}))

Simplify

- 3 (- cos (\frac{1 - x}{3})) : 3 cos (\frac{1}{3} (1 - x))

= 3 cos (\frac{1}{3} (1 - x))

Add a constant to the solution

= 3 cos (\frac{1}{3} (1 - x)) + C

cos^{2} (ln (e^{2 x + 1} (- 2 x)))

\frac{3 e ^{- π s}}{s ^{2} + 9}

f (x) = 4 x^{2} - 1, (- 6, 3)

y^{^{'}} = (x + y - 2)^{2}, y (0) = 0

2 cos (π t)

What is the integral of sin((1-x)/3) ?
The integral of sin((1-x)/3) is 3cos(1/3 (1-x))+C