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Question Number 53259 by Tawa1 last updated on 19/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$I={\int }_0^{\pi }\frac{xdx}{a^2{cos}^{2}x+b^2{sin}^{2}x}dx$$$$I={\int }_0^{\pi }\frac{(\pi -x)dx}{a^2{cos}^2(\pi -x)+b^2{sin}^2(\pi -x)}$$$$2I={\int }_0^{\pi }\frac{x+\pi -x}{a^2{cos}^{2}x+b^2{sin}^{2}x}dx$$$$\frac{2I}{\pi }={\stackrel{\pi }{\int }}_0\frac{dx}{{cos}^{2}x(a^2+b^2{tan}^{2}x)}$$$$={\int }_0^{\pi }\frac{{sec}^{2}xdx}{b^2(\frac{a^2}{b^2}+{tan}^{2}x)}$$$$\frac{2I}{\pi }=\frac{1}{b^2}{\int }_0^{\pi }\frac{{sec}^{2}x}{\frac{a^2}{b^2}+{tan}^{2}x}dx$$$$wait...$$$$f\left(x\right)=\frac{{sec}^{2}x}{\frac{a^2}{b^2}+{tan}^{2}x}$$$$f(\pi -x)=\frac{{sec}^2(\pi -x)}{\frac{a^2}{b^2}+{tan}^2(\pi -x)}=f\left(x\right)$$$$so$$$$\frac{2I}{\pi }=\frac{1}{b^2}\times 2{\int }_0^{\frac{\pi }{2}}\frac{{sec}^{2}x}{\frac{a^2}{b^2}+{tan}^{2}x}$$$$\frac{{Ib}^2}{\pi }={\int }_0^{\frac{\pi }{2}}\frac{{sec}^{2}x}{\frac{a^2}{b^2}+{tan}^{2}x}dx$$$$letk=tanx[dk={sec}^{2}xdx$$$$\frac{{Ib}^2}{\pi }={\int }_0^{\mathrm{\infty }}\frac{dk}{\frac{a^2}{b^2}+k^2}$$$$\frac{{Ib}^2}{\pi }=\frac{1}{\frac{a}{b}}\times \mid {tan}^{-1}\left(\frac{k}{\frac{a}{b}}\right){\mid }_0^{\mathrm{\infty }}$$$$I=\frac{\pi }{ab}[{tan}^{-1}\left(\mathrm{\infty }\right)-{tan}^{-1}\left(0\right)]$$$$I=\frac{\pi }{ab}\times \frac{\pi }{2}=\frac{{\pi }^2}{2ab}$$

Commented by Tawa1 last updated on 19/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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