Calculus - integration

62

Hint:

One way which is most often used is to find

integral of âˆštanx+âˆšcotx

and of âˆštanx-âˆšcotx

262

Aditya Bhutra ·2011-06-19 02:20:22

let tanx = z² , sec²x dx = 2z dz

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

= âˆ«{z²+11+z⁴ } + âˆ«{z²-11+z⁴ }

= âˆ«{z^-2+1z²+z^-2 } +âˆ«{-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 = âˆš2.tan^-1 (t/âˆš2)

2nd frac. -

dpp²-2 = (1/2âˆš2 ).ln(|p-2p+2 |)

substituing for x we get the final ans.

1

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

62

Good work aditya..

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

integration