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        ∫(1+ tan x)1/2dx
8 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans:$I = \int \sqrt{1+tanx}dx$$1+tanx = t^{2}$$sec^{2}x.dx = 2t.dt$$dx = \frac{2t}{(t^{2}-1)^{2}+1}dt$$I = \int \frac{2t^{2}}{(t^{2}-1)^{2}+1}dt$Simply apply the partial fraction rule here.$I = 2[\frac{tan^{-1}(\frac{t}{\sqrt{-1-i}})}{(-1-i)^{3/2}}-\frac{tan^{-1}(\frac{t}{\sqrt{-1+i}})}{(-1+i)^{3/2}}] + constant$$I = 2[\frac{tan^{-1}(\frac{\sqrt{1+tanx}}{\sqrt{-1-i}})}{(-1-i)^{3/2}}-\frac{tan^{-1}(\frac{\sqrt{1+tanx}}{\sqrt{-1+i}})}{(-1+i)^{3/2}}] + constant$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
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