Question

integration of:
sqrt( (1+x)/(1-x) ) I tried using x=cos x but the answer seems to be diff.

Answer 1

I got the answer as:
-cos inverse (x) - sin ( cos inverse (x) ) + c which is wrong. The correct answer is: sin inverse (x) - sqrt(1-x^2) + C

Answer 2

try x=cos 2y
what you get after simplifying the square root ?

Answer 3

still no use, since the first term of the answer is arcsin(x) , i think either the answer given is wrong or we have to take t=sinx ,
with t=sinx i dont know how to do?

Answer 4

try integral by u-subs
let u=1-x --------> x=1-u
du = dx

= int(1+x)/(1-x) dx
= int (1+1-u)/u du
= int (2-u)/u du
= int 2/u du - int 1 du
= ...

Answer 5

what about sqrt(...) :P

Answer 6

haha, misread :D

Answer 7

\[\int\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1-x}{1-x}}dx\\
\int\frac{\sqrt{1-x^2}}{1-x}dx\]
\[u=1-x\Rightarrow x=1-u\\
du=-dx\]
\[-\int\frac{\sqrt{1-(1-u)^2}}{u}du\]
\[1-u=\sin t\\
-du=\cos t\;dt\]
\[\int\frac{\sqrt{1-\sin^2t}}{1-\sin t}\cos t\;dt\\
\int\frac{\cos^2t}{1-\sin t}dt\\
\int\frac{1-\sin^2t}{1-\sin t}dt\\
\int\frac{(1+\sin t)(1-\sin t)}{1-\sin t}dt\\
\int(1+\sin t)\;dt\\
t-\cos t+C\]
And make sure to sub back.

Answer 8

brilliant. thanks a lot