Question

the value of definite integral In(pi/2-t^2/2)dt is

Answer 1

Answer:

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Answer 2

Given : In(pi/2-t^2/2)dt

We solve the given function in following way;

$\int_{0}^{1} \log \left(\frac{\pi}{2}-\frac{t^{2}}{2}\right) d x=\log \left(\frac{1}{2}\left(\pi-t^{2}\right)\right)$

$\text { Series expansion of the integral at } \mathrm{t}=0 \text { : }\\\log \left(\frac{\pi}{2}\right)-\frac{t^{2}}{\pi}+O\left(t^{4}\right)$${ Series expansion of the intearal at } \mathrm{t}=\text {-sart(?) }\\\left(\log (t+\sqrt{\pi})+\frac{\log (\pi)}{2}\right)-\frac{t+\sqrt{\pi}}{2 \sqrt{\pi}}-\frac{(t+\sqrt{\pi})^{2}}{8 \pi}-\frac{(t+\sqrt{\pi})^{3}}{24 \pi^{3 / 2}}+O\left((t+\sqrt{\pi})^{4}\right)$

$\text { Series expansion of the integral at } \mathrm{t}=\text { sqrt(?): }\\\left\{\begin{array}{c}\left(\log (-\sqrt{\pi}(t-\sqrt{\pi}))+\frac{t-\sqrt{\pi}}{2 \sqrt{\pi}}-\right. \\\left.\frac{(t-\sqrt{\pi})^{2}}{8 \pi}+O\left((t-\sqrt{\pi})^{3}\right)\right)^{*} \\\log (-\sqrt{\pi}(t-\sqrt{\pi}))+ \\\frac{t-\sqrt{\pi}}{2 \sqrt{\pi}}-\frac{(t-\sqrt{\pi})^{2}}{8 \pi}+O\left((t-\sqrt{\pi})^{3}\right)\end{array}\right.$

$\int \log \left(\frac{\pi}{2}-\frac{t^{2}}{2}\right) d x=x \log \left(\frac{\pi}{2}-\frac{t^{2}}{2}\right)+\text { constant }$

The value of definite integral is;

$\int \log \left(\frac{\pi}{2}-\frac{t^{2}}{2}\right) d x=x \log \left(\frac{\pi}{2}-\frac{t^{2}}{2}\right)+\text { constant }$