I use the substitution x^2=z.With this substitution the integrand becomes cosz/2z^(1/2).Is it possible to integrate w.r.t. z.Please help me.
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2Are there limits in your integal? (If so, search for Fresnel integrals). If you are looking for an antiderivtive, it is not an elementary function. – mrf Oct 22 '15 at 05:55
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No,this is an indefinite integration. – Arnab Chattopadhyay Oct 22 '15 at 05:57
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I feel like I should ask, are you sure it is $\cos(x^2)$ and not $\cos^2(x)$? – CPM Oct 22 '15 at 06:01
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Yes you have rightly thought. – Arnab Chattopadhyay Oct 22 '15 at 06:03
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1http://math.stackexchange.com/questions/1239046/evaluate-int-sinx2-mathrmdx – ahorn Oct 22 '15 at 06:06
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I still dont find my answer.Is the integrand possible to integrate? – Arnab Chattopadhyay Oct 22 '15 at 06:33
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1Yes, it is but the result involves Fresnel integrals which cannot express in terms of elementary functions. – Claude Leibovici Oct 22 '15 at 06:34
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As said, $$\int \cos(x^2)\ dx=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} x\right)+K$$ where appears Fresnel integral. This cannot be expressed in terms of elementary functions.
What you could do is to use series starting from $$\cos(y)=\sum_{n=0}^\infty \frac{(-1)^n y^{2n}}{(2n)!}$$ which makes $$\cos(x^2)=\sum_{n=0}^\infty \frac{(-1)^n x^{4n}}{(2n)!}$$ and, by integration $$\int \cos(x^2)\ dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+1}}{(4n+1)(2n)!}+K$$
Edit
Suppose that you need to compute $$I=\int_0^1 \cos(x^2)\ dx$$ Using the above summation for $p$ terms, you would get $$I_{0}=1.000000$$ $$I_{1}=0.900000$$ $$I_{2}=0.904630$$ $$I_{3}=0.904523$$ $$I_{4}=0.904524$$
Claude Leibovici
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