integration

i am not able to solve this sum,someone help

∫√tanx dx = ?

4 Answers

62
Lokesh Verma ·

Hint:

One way which is most often used is to find

integral of √tanx+√cotx

and of √tanx-√cotx

262
Aditya Bhutra ·

let tanx = z2 , sec2x dx = 2z dz

therefore I = ∫(z. 2z dz)/sec2x
I= ∫2z21+z4 dz

= ∫{z2+11+z4 } + ∫{z2-11+z4 }

= ∫{z-2+1z2+z-2 } +∫{-z-2+1z2+z-2 }

now,
z2 +z-2 = (z-1/z)2 +2 (for first fraction)
z2 +z-2 = (z+1/z)2 -2 (for 2nd frac.)

replace in previous step then
take z-1/z =t for first frac. and
z+1/z = p for 2nd frac.

then just standard integrals

1st frac. -

dtt2 +2 = √2.tan-1 (t/√2)

2nd frac. -

dpp2-2 = (1/2√2 ).ln(|p-2p+2 |)

substituing for x we get the final ans.

1
Debosmit Majumder ·

thnx a lot sir,i`ve solved it by your method.

62
Lokesh Verma ·

Good work aditya..

@Debosmit.. Good to see u getting regular here.

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