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Question Number 139002 by BHOOPENDRA last updated on 21/Apr/21

Commented by Dwaipayan Shikari last updated on 22/Apr/21

$$Probability{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mid \mathrm{\Psi }\left(x\right){\mid }^{2}dx=1$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{A}^2{cos}^{4}xdx=A^2\mathrm{\Gamma }\left(\frac{5}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)=A^2\frac{3\pi }{4}$$$$\Rightarrow 1=A^2\frac{3\pi }{4}\Rightarrow A=\pm \frac{2}{\sqrt{3\pi }}$$

Commented by Dwaipayan Shikari last updated on 22/Apr/21

$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{A}^2{cos}^{4}xdx=1$$$$={\int }_0^{\frac{\pi }{2}}{A}^2{cos}^{4}xdx=\frac{1}{2}\Rightarrow {\int }_0^{\frac{\pi }{4}}{A}^2{cos}^{4}xdx=\frac{1}{4}or25\mathrm{\%}$$$$Probability=\frac{1}{4}$$

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