calculus - How to compute the integral of $int sin(1-x^2),dx$

I tried substitution, $u=1-x^2,\, du = -2x\,dx ,\, dx = -\frac{du}{2x}$, yielding:

\begin{align} & \int \sin u \left(-\frac{du} 2 \right) \\[10pt] = {} & -\frac{1}{2}\int \sin(u)\,du \\[10pt] = {} & -\frac{1}{2}\cos u \\[10pt] = {} & -\frac{1}{2}\cos (1-x^2) \end{align}

WolframAlpha is showing something completely different, however. What's wrong with my solution?

Answer

Your substitution is wrong.

Set $$1 - x^2 = a ~~~~~~~ dx = \frac{-1}{2 \sqrt{1 - a}} da$$

Hence

$$-\frac{1}{2}\int \frac{\sin(a)\ da}{\sqrt{1 - a}}$$

Elementary knowledge of analysis tell you that this is a Fresnel Type integral, and the solution can be written as

$$-\frac{1}{2}\sqrt{2 \pi }\cos(1) \left( S\left(\frac{\sqrt{2-2 x}}{\sqrt{\pi }}\right)-\cot(1) C\left(\frac{\sqrt{2-2 x}}{\sqrt{\pi }}\right)\right)$$

Where $S$ and $C$ stand for "Fresnel Sine and Cosine Integral"

Cot stands for the co-tangent.

More on Fresnel Integrals:

https://en.wikipedia.org/wiki/Fresnel_integral