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what is the integral of(1-tan 2 x) 1/2 dx???

what is the integral of(1-tan2x)1/2dx???

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
I = \int \sqrt{1-tan^{2}x}dx
t = tanx
dt = sec^{2}x.dx
dx = \frac{dt}{1+t^{2}}
I = \int \frac{\sqrt{1-t^{2}}}{1+t^{2}}dt
t = sin(s)
dt = cos(s).ds
I = \int \frac{cos^{2}(s)}{1+sin^{2}(s)}ds
I = \int \frac{sec^{2}(s)}{sec^{4}(s)+sec^{2}(s).tan^{2}(s)}ds
u = tan(s)
du = sec^{2}(s).ds
I = \int \frac{1}{2u^{4}+3u^{2}+1}du
Using the partial fraction rule, we have
I = \sqrt{2}tan^{-1}(\sqrt{2}u)-tan^{-1}(u)+constant
I = \sqrt{2}tan^{-1}(\sqrt{2}tan(s))-s+constant
I = \sqrt{2}tan^{-1}(\frac{\sqrt{2}t}{\sqrt{1-t^{2}}})-sin^{-1}(t)+constant
I = \sqrt{2}tan^{-1}(\frac{\sqrt{2}tan(x)}{\sqrt{1-tan^{2}(x)}})-sin^{-1}(tan(x))+constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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