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Integral Calculus

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 Joined: 20 Jun 2012 Post: 3
20 Jun 2012 09:17:20 IST
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Integral (x+(cos inverse 3x)^2 )/ (root of(1-9x^2))dx
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Integral Calculus

=A root of(1-9x^2)+b(cos inverse 3x)^3 +c,,Find A,B?

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Joined: 19 Jan 2008
Posts: 1070
26 Jun 2012 14:52:49 IST
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$\hspace{-16}\bf{\int\frac{x+\left(\cos^{-1}(3x)\right)^2}{\sqrt{1-9x^2}}dx}\\\\\\ Now Let \bf{x=\frac{1}{3}\cos t\Leftrightarrow dx =-\frac{1}{3}\sin tdt}\\\\\\ \bf{\int\frac{\frac{1}{3}\cos t+t^2}{\sin t}\times -\frac{1}{3}\sin tdt}\\\\\\ \bf{=-\frac{1}{9}\int \cos tdt-\frac{1}{3}\int t^2dt}\\\\\\ \bf{=-\frac{1}{9}\sin t-\frac{1}{9}.t^3+C}\\\\\\ So \bf{\int\frac{x+\left(\cos^{-1}(3x)\right)^2}{\sqrt{1-9x^2}}dx=-\frac{1}{9}\sqrt{1-9x^2}-\frac{1}{9}.\left(\cos^{-1}\;3x\right)^3+C}\\\\\\ So Camparing, We Get\\\\\\ \bf{A=-\frac{1}{9}\;\;\;\;,B=-\frac{1}{9}}$

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27 Jun 2012 19:13:52 IST
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Thank you

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28 Jun 2012 14:57:45 IST
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Thanks

New kid on the Block

Joined: 20 Jun 2012
Posts: 3
1 Jul 2012 19:53:22 IST
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thankyou........

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