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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: definite integrals
Forum Index -> Integral Calculus like the article? email it to a friend.  
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ankitagg (302)

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1..


2..   (whole is in multiplication with x)


3...


 

    
thedumbheadwithnobrain (887)

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I=\int_{0}^{\pi}\frac{xtanx}{secx+tanx}dx\\I=\int_{0}^{\pi}xtanx(secx-tanx)dx\\I=\int_{0}^{\pi}x\;tanx\;secx\;dx-\int_{0}^{\pi}x\;sec^2x\;dx+\int_{0}^{\pi}x\;dx\\I=I_1+I_2+I_3\\I_1=\int_{0}^{\pi}(\pi-x)tanx\;secx\;dx\\I_1=\frac{\pi}{2}\int_{0}^{\pi}secx\;tanx\;dx=\frac{\pi}{2}[secx]_{0\;to\;\pi}=-\pi\\I_2=\int_{0}^{\pi}(\pi-x)sec^2x\;dx\\I_2=\frac{\pi}{2}[tanx]_{0\;to\;\pi}=0\\I_3=\int_{0}^{\pi}x\;dx=\frac{\pi^2}{2}\\I=I_1+I_2+I_3\\I=\frac{\pi^2}{2}-\pi

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thedumbheadwithnobrain (887)

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I=\int_{0}^{\pi}\frac{x}{1+cos^2x}\;dx\\I=\frac{\pi}{2}\int_{0}^{\pi}\frac{dx}{1+cos^2x}\\I=\frac{\pi}{2}\int_{0}^{\pi}\frac{sec^2x}{sec^2x+1}dx\\I=\pi\int_{0}^{\frac{\pi}{2}}\frac{sec^2x\;dx}{tan^2x+2}\\I=\left[\frac{\pi}{\sqrt{2}}\left(tan^{-1}\left(\frac{tanx}{\sqrt{2}}\right)\right)\right]_{0\;to\;\frac{\pi}{2}}\\I=\frac{\pi}{\sqrt{2}}\left(\frac{\pi}{2}-0\right)\\I=\frac{\pi^2}{2\sqrt{2}}

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thedumbheadwithnobrain (887)

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I=\int_{0}^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)\left(\sqrt{tanx}+\frac{1}{\sqrt{tanx}}\right)\;dx\\I=\frac{\pi}{4}\int_{0}^{\frac{\pi}{2}}\left(\sqrt{tanx}+\frac{1}{\sqrt{tanx}}\right)\;dx\\Put\;tanx=t^2,sec^2x\;dx=2t\;dt,\;dx=\frac{2t}{1+t^4}\\I=\frac{\pi}{2}\int_{0}^{\infty}\frac{t^2+1}{t^4+1}dt\\I=\frac{\pi}{2}\int_{0}^{\infty}\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt\\I=\frac{\pi}{2\sqrt{2}}tan^{-1}(\frac{t-\frac{1}{t}}{\sqrt{2}})\\I=\frac{\pi}{2\sqrt{2}}\left[tan^{-1}(\infty)-tan^{-1}(-\infty)\right]\\I=\frac{\pi}{2\sqrt{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\I=\frac{\pi^2}{2\sqrt{2}}

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ankitagg (302)

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THE ANSWERS ARE


1. 1/2  PIE(PIE-2)


2.(PIE)^2/2*(ROOT)2


3..IN THIRD QUESTION IT IS COS^2X NOT COS2X


ANS.. (PIE)^2/2*(ROOT)2

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