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Integral Calculus> integral of root tanx divided by only sec...

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integral of root tanx divided by only secx

Jamal , 10 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$I = \int \frac{\sqrt{tanx}}{secx}dx$

$I = \int \sqrt{sinx.cosx}dx$

$I = \frac{1}{\sqrt{2}}\int \sqrt{sin2x}dx$

$z = cos2x + isin2x = e^{2ix}$

$dz = 2ie^{2ix}dx = 2izdx$

$\frac{1}{z} = cos2x - isin2x = e^{-2ix}$

$z-\frac{1}{z} = 2isin2x$

$sin2x = \frac{1}{2i}(z-\frac{1}{z})$

$I = \frac{1}{\sqrt{2}}\int \sqrt{\frac{1}{2i}(z-\frac{1}{z})}\frac{dz}{2iz}$

$I = \frac{1}{4i^{3/2}}\int \sqrt{(z-\frac{1}{z})}\frac{dz}{z}$

$I = \frac{1}{4i^{3/2}}.(2\sqrt{z-\frac{1}{z}}(\frac{F(sin^{-1}(\sqrt{z})|-1)-2\sqrt{z}E(sin^{-1}(\sqrt{z})|-1)}{\sqrt{1-z^{2}}}-1))+constant$ $I = (\frac{1}{4i^{3/2}}.(2\sqrt{z-\frac{1}{z}}(\frac{F(sin^{-1}(\sqrt{z})|-1)-2\sqrt{z}E(sin^{-1}(\sqrt{z})|-1)}{\sqrt{1-z^{2}}}-1)))+constant$

Thanks & Regards

Jitender Singh

IIT Delhi

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