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?

  • 1
    $\begingroup$ $dx=-\frac{du}{2x}$ is correct, but you didn't use this in the next line $\endgroup$ – A. Goodier Dec 17 '17 at 18:21
  • $\begingroup$ @woofy I just factored it, shouldn't we do that with constants? $\endgroup$ – Trey Dec 17 '17 at 18:21
  • 1
    $\begingroup$ Where did the $x$ go? $\endgroup$ – A. Goodier Dec 17 '17 at 18:22
  • $\begingroup$ $$\int{\sin{(1-x^{2})}}dx=\sin{(1)}\int\cos{(x^{2})}dx+\cos{(1)}\int\sin{(x^{2})}dx$$ Now you can use the fact that $$\int{e^{ix^{n}}dx}=x_{1}F_{1}\Big(\frac{1}{n}, 1+\frac{1}{n}, ix^{n}\Big)$$ $\endgroup$ – Kiryl Pesotski Dec 17 '17 at 18:29
  • $\begingroup$ @woofy can I factor $\frac{1}{2}*x$ then? $\endgroup$ – Trey Dec 17 '17 at 18:32

Your substitution is wrong.

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


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

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:


  • $\begingroup$ How did $\frac{-1}{2 \sqrt{1 - a}} da$ become $-\frac{1}{2}\int \frac{\sin(a)\ da}{\sqrt{1 - a}}$? $\endgroup$ – Trey Dec 17 '17 at 19:26
  • $\begingroup$ Integration by substitution. You had a sine before, you have to maintain the sine after. Also the $-1/2$ constant has been taken out $\endgroup$ – Von Neumann Dec 17 '17 at 19:49


$=\sin1\int\cos x^2~dx-\cos1\int\sin x^2~dx$

$=\sqrt{\dfrac{\pi}{2}}\left(\sin1~C\left(\sqrt{\dfrac{2}{\pi}}x\right)-\cos1~S\left(\sqrt{\dfrac{2}{\pi}}x\right)\right)+C$ (according to http://mathworld.wolfram.com/FresnelIntegrals.html)


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