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integral of root tanx divided by only secx

integral of root tanx divided by only secx

Grade:

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 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))+constantI = (\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
askIITians Faculty

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