What is the integral of cos^3(pix-1) ?

The integral of cos^3(pix-1) is 1/pi (sin(pix-1)-(sin^3(pix-1))/3)+C

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integral of cos^3(pix-1)

\int cos^{3} (π x - 1) d x

\frac{1}{π} (sin (π x - 1) - \frac{sin ^{3} ( π x - 1 )}{3}) + C

Solution steps

\int cos^{3} (π x - 1) d x

Apply u-substitution

: \int cos^{3} (u) \frac{1}{π} d u

= \int cos^{3} (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int cos^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int 1 - v^{2} d v

= \frac{1}{π} \cdot \int 1 - v^{2} d v

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}{π} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{π} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{π} (sin (π x - 1) - \frac{sin ^{3} ( π x - 1 )}{3})

Add a constant to the solution

= \frac{1}{π} (sin (π x - 1) - \frac{sin ^{3} ( π x - 1 )}{3}) + C

f (x) = \frac{1}{2 - 3 x}

\frac{d}{d x} ((\frac{x ^{2} - 9}{4 x + 5})^{4})

\frac{1}{x} \frac{d y}{d x} - \frac{2}{x ^{2}} y = x^{2} e^{- x}

\frac{5}{x ^{5}} + \frac{3}{x ^{2}}

\int \frac{y - y}{y ^{2}} d y

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