What is the integral of cos^2((4-x)/3) ?

The integral of cos^2((4-x)/3) is -1/2 (4-x+3/2 sin((2(4-x))/3))+C

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integral of cos^2((4-x)/3)

\int cos^{2} (\frac{4 - x}{3}) d x

- \frac{1}{2} (4 - x + \frac{3}{2} sin (\frac{2 ( 4 - x )}{3})) + C

Solution steps

\int cos^{2} (\frac{4 - x}{3}) d x

Apply u-substitution

: \int - cos^{2} (\frac{u}{3}) d u

= \int - cos^{2} (\frac{u}{3}) d u

Take the constant out:

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

= - \int cos^{2} (\frac{u}{3}) d u

Rewrite using trig identities

= - \int \frac{1}{2} (1 + cos (\frac{2 u}{3})) 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 1 + cos (\frac{2 u}{3}) 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} (\int 1 d u + \int cos (\frac{2 u}{3}) d u)

\int 1 d u = u

\int cos (\frac{2 u}{3}) d u = \frac{3}{2} sin (\frac{2 u}{3})

= - \frac{1}{2} (u + \frac{3}{2} sin (\frac{2 u}{3}))

Substitute back

u = 4 - x

= - \frac{1}{2} (4 - x + \frac{3}{2} sin (\frac{2 ( 4 - x )}{3}))

Add a constant to the solution

= - \frac{1}{2} (4 - x + \frac{3}{2} sin (\frac{2 ( 4 - x )}{3})) + C

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What is the integral of cos^2((4-x)/3) ?
The integral of cos^2((4-x)/3) is -1/2 (4-x+3/2 sin((2(4-x))/3))+C