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Integral Calculus

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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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jagdish singh

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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}}$$

Mohd Sufiyan

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27 Jun 2012 19:13:52 IST

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Thank you

yogesh

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28 Jun 2012 14:57:45 IST

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Thanks

akhilraj

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1 Jul 2012 19:53:22 IST

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thankyou........

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