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Find the integral of x sin x2 cos x2

Evaluate the following integral:

    \[ \int x \sin x^2 \cos x^2 \, dx. \]


First, we make the substitution

    \[ u = x^2 \qquad \implies \qquad du = 2x \, dx. \]

Giving us

    \[ \int x \sin (x^2) \cos (x^2) \, dx &= \frac{1}{2} \int \sin u \cos u \, du. \]

Then we recall the trig identity \sin (2u) = 2 \sin u \cos u to obtain,

    \[ \frac{1}{2} \int \sin u \cos u \, du = \frac{1}{4} \int \sin (2u) \, du. \]

We make another substitution,

    \[ t = 2u \qquad \implies \qquad dt = 2 \, du \]

to arrive at

    \begin{align*}   \frac{1}{4} \int \sin (2u) \, du &= \frac{1}{8} \int \sin t \, dt \\  &= \frac{1}{8} (-\cos t) + C \\  &= - \frac{\cos (2u)}{8} + C. \end{align*}

Now we recall the other double angle trig identity, \cos (2u) = 1 - 2 \sin^2 u so we have

    \begin{align*}  -\frac{\cos (2u)}{8} + C &= \frac{2 \sin^2 u - 1}{8} + C \\  &= \frac{\sin^2 u}{4} - \frac{1}{8} + C \\  &= \frac{\sin^2 u}{4} + C &(\text{absorbing the } \frac{1}{8} \text{ into } C)\\  &= \frac{1}{4}\sin^2 x^2 + C. \end{align*}

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