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Find the integral of sin (x-1)(1/4)

by RoRi

Evaluate the following integral:

    \[ \int \sin \sqrt[4]{x-1} \, dx. \]

This will take several steps. First we observe the derivative

    \[ D\left( (x-1)^{\frac{1}{4}} \right) = \frac{1}{4(x-1)^{\frac{3}{4}}}. \]

Then we multiply by 1 to get this derivative into the integrand,

    \[ \int \sin (x-1)^{\frac{1}{4}} \, dx = \int \left(\left(4(x-1)^{\frac{3}{4}} \right) \left( \frac{1}{4(x-1)^{\frac{3}{4}}} \cdot \sin (x-1)^{\frac{1}{4}} \right)\right) \, dx. \]

Now we make a substitution,

    \[ t = (x-1)^{\frac{1}{4}} \qquad \implies \qquad dt = \frac{1}{4 (x-1)^{\frac{3}{4}}} \, dx. \]

Giving us,

    \[ \int \sin (x-1)^{\frac{1}{4}} \, dx = \int 4t^3 \sin t \, dt. \]

Now we recall this previous exercise where we used integration by parts (a couple of times) to find

    \[ \int x^3 \sin x \, dx = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C\]

Therefore we have,

    \begin{align*} \int 4t^3 \sin t \, dt &= 4 \int t^3 \sin t \, dt \\ &= 4 (-t^3 \cos t + 3t^2 \sin t + 6t \cos t - 6 \sin t) + C. \end{align*}

Now, since t = (x-1)^{\frac{1}{4}} we have

    \begin{align*} &4 ( -(x-1)^{\frac{3}{4}} \cos (x-1)^{\frac{1}{4}} + 3(x-1)^{\frac{1}{2}} \sin (x-1)^{\frac{1}{4}} + 6(x-1)^{\frac{1}{4}} \cos (x-1)^{\frac{1}{4}} - 6 \sin (x-1)^{\frac{1}{4}} ) + C \\[10pt] &= \left( 12(x-1)^{\frac{1}{2}} - 24\right)\sin (x-1)^{\frac{1}{4}} -4 \left((x-1)^{\frac{3}{4}} - 6(x-1)^{\frac{1}{4}} \right) \cos (x-1)^{\frac{1}{4}} + C \end{align*}

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