IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

IITPATIENT

Integrate (cos2x)^1/2 / sinx

deedee

hey quite nice sum i did it 2day itself .wait postin ..........it was asked in sum iit paper for 6 mrks ....

deedee

m just givin u d steps u do ...

first rite it as (cot^2x-1)^1/2 then substitute cotx=t

then ur int bcums -(1-t^2)^1/2 /(t^2+1)

then write it as 2/(t^2+1)*(t^2-1)^1/2 -1/(t^2-1)^1/2

nw d secodn part u can int easily ffor d first part

t=secy

nw ur int reduces to 2secydy/(sec^2y+1)

noe write it as 2cosy/(1+cos^2y)

cos^2y-->z

so nw ur int bcums

dz/(z)*(2-z)^1/2

nw put 2-z--->u

so ur int bcums 2du/(2-u^2) u can int this easily

shorter emthods r welcumed.......... :)

rudra.panda

An attempt!

$I=\int \frac{\sqrt{\cos 2x}}{\sin x}\ dx$

$=\int \frac{\cos 2x}{\sin x\sqrt{\cos 2x}}\ dx$

$=\int \frac{1-2\sin^2 x}{\sin x\sqrt{\cos 2x}}\ dx$

$=\int \frac{1}{\sin x\sqrt{\cos^2x-\sin^2 x}}\cdot dx-2\int\frac{\sin x}{\sqrt{2\cos^2 x-1}}\ dx$

$=\int \frac{\csc^2 x}{\sqrt{\cot^2 x-1}}\cdot dx-\frac{2}{\sqrt{2}}\int\frac{\sin x}{\sqrt{\cos^2 x-(1/2)}}\ dx$

$=-\int \frac{dt}{\sqrt{t^2-1}}-\sqrt{2}\int\frac{-ds}{\sqrt{s^2-(1/2)^2}}$ where t=cot x and s=cos x

$=-\log|t+\sqrt{t^2-1}|+\sqrt{2}\log|s+\sqrt{s^2-/2}|+c$

$=-\log|\cot x+\sqrt{\cot^2 x-1}|+\sqrt{2}\log\left|\cos x+\sqrt{\cos^2 x-\frac{1}{2}}\right|+c$

I think this is correct any more methods?

ailaSachin

Done by expert puneet :

Mirka

In above Rudra's method,
in the 5th step,
shudn't it be cosec x and not cosec^2 x???