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Grade: 12
        
What is the solution for ∫√(1 + tanx) dx? 
Please give me answer
 
 
7 months ago

Answers : (1)

Aditya Gupta
1717 Points
							
put y^2=1 + tanx
2ydy= sec^2xdx=(1 + tan^2x)dx=(1 + (y^2 – 1)^2)dx
so that I= ∫2y^2dy/(1 + (y^2 – 1)^2)
= ∫2y^2dy/(y^4 – 2y^2+2)
now this last integral is superbly tricky.
write 2y^2/(y^4 – 2y^2+2) as 2/(y^2 – 2+2/y^2)
= (1 – 2/y^2)/(y^2 – 2+2/y^2) + (1+2/y^2)/(y^2 – 2+2/y^2)
this way we have two integrals. in the first integral, write the denominator y^2 – 2+2/y^2 as (y+2/y)^2 – 2(1+2) and then substitute y+2/y= z and the integral will turn into a standard partial fraction form.
in the second interal, write the denominator y^2 – 2+2/y^2 as (y – 2/y)^2 – 2+22 and then substitute y – 2/y= t and the integral will turn into a standard partial fraction form.
7 months ago
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