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Integral Calculus

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4 Feb 2012 07:34:06 IST

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integration of (sin2x)/a^2sin^2x+b^2cos^2x

Mathematics

integration of (sin2x)/a²sin²x+b²cos²x..am not getting its ans plz guide me ...also another same integral ....sin2x/(a+bcosx)^{2...guide me on this too}

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jagdish singh

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4 Feb 2012 12:11:26 IST

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$\hspace{-16}\int\frac{\sin 2x}{a^2\sin^2 x+b^2\cos^2 x}dx$\\\\\\ Put $a^2\sin^2x+b^2\cos^2x=t\Leftrightarrow (a^2-b^2)\sin 2xdx=dt$\\\\\\ $\frac{1}{(a^2-b^2)}\int\frac{1}{t}dt=\frac{1}{a^2-b^2}\ln \mid t \mid+C$\\\\\\ $=\frac{1}{a^2-b^2}\ln \left|a^2\sin^2x+b^2\cos^2x\right|+C$$

jagdish singh

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5 Feb 2012 15:45:10 IST

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$\hspace{-16}\int\frac{\sin 2x}{(a+b\cos x)^2}dx$\\\\\\ $\int \frac{2\sin x.\cos x}{(a+b\cos x)^2}dx$\\\\\\ Put $a+b\cos x=t\Leftrightarrow \sin xdx =-\frac{1}{b}dt$\\\\\\ $-\frac{2}{b^2}\int\frac{(t-a)}{t^2}dt=-\frac{2}{b^2}\int\frac{1}{t}dt+\frac{2a}{b^2}\int\frac{1}{t^2}dt$\\\\\\ $-\frac{2}{b}\ln \mid t \mid -\frac{2a}{b^2}.\frac{1}{t}+C$\\\\\\ So $\int\frac{\sin 2x}{(a+b\cos x)^2}dx=-\frac{2}{b}\ln\left| a+b\cos x\left| -\frac{2a}{b^2}.\frac{1}{a+b\cos x}+C$$

alok vats

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7 Feb 2012 07:49:59 IST

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Thanks Bro ...Thanking you in anticipation

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