How to compute the integral of ∫sin⁡(1−x2)dx I tried substitution, u=1−x2, du=−2xdx, dx=−du2x, yielding:∫sin⁡u(−du2)=−12∫sin⁡(u)du=−12cos⁡u=−12cos⁡(1−x2)WolframAlpha is showing something completely different, however. What's wrong with my solution?

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How to compute the integral of ∫sin⁡(1−x2)dx I tried substitution, u=1−x2, du=−2xdx, dx=−du2x, yielding:∫sin⁡u(−du2)=−12∫sin⁡(u)du=−12cos⁡u=−12cos⁡(1−x2)WolframAlpha is showing something completely different, however. What&#039;s wrong with my solution?

utloverej · Accepted Answer

Your substitution is wrong.Set1−x2=a    dx=−121−adaHence−12∫sin⁡(a) da1−aElementary knowledge of analysis tell you that this is a Fresnel Type integral, and the solution can be written as−122πcos⁡(1)(S(2−2xπ)−cot⁡(1)C(2−2xπ))Where S and C stand for &quot;Fresnel Sine and Cosine Integral&quot;

How to compute the integral of \int \sin(1-x^2) dx I

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