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Question Number 211908 by Spillover last updated on 23/Sep/24

Answered by BHOOPENDRA last updated on 24/Sep/24

$$I={\int }_0^{\sqrt{3}}\left(\underset{k=0}{\sum ^{\mathrm{\infty }}}\sin \left(\frac{k\pi }{2}\right)x^k\right)dx$$$$Thepatternfor\sin \left(k\pi /2\right)$$$$fork=0$$$$\sin \left(\frac{k\pi }{2}\right)=\sin \left(0\right)=0$$$$Fork=1$$$$\sin \left(1\times \frac{\pi }{2}\right)=\sin \left(\pi /2\right)=1$$$$Fork=2$$$$\sin \left(2\times \frac{\pi }{2}\right)=\sin \left(\pi \right)=0$$$$Fork=3$$$$\sin \left(\frac{3\pi }{2}\right)=-1,Fork=4\sin \left(\frac{4\pi }{2}\right)=0$$$$Fork=5$$$$\sin \left(\frac{5\pi }{2}\right)=1$$$$S\left(x\right)=(x-x^3+x^5-x^7+x^9+..........)$$$$S\left(x\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\sin \left(\frac{k\pi }{2}\right)x^k=\frac{x}{(1+x^2)}$$$$I={\int }_0^{\sqrt{3}}S\left(x\right)dx$$$$={\int }_0^{\sqrt{3}}\frac{x}{1+x^2}dx$$$${\left[ln(1+x^2)/2\right]}_0^{\sqrt{3}}$$$$=\frac{ln\left(4\right)}{2}=ln2$$

Commented by Spillover last updated on 24/Sep/24

$$good.thanks$$

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