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suchintan mondal Grade: 12
        

∫(1+ tan x)1/2dx

7 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
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 & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
2 years ago
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