∫sin-1x1-x2dx=
logsin-1x+c
12sin-1x2+c
log1-x2+c
sincos-1x+c
Explanation for the correct option:
Evaluate : ∫sin-1x1-x2dx
consider the given Equation as
I=∫sin-1x1-x2dx
Let us assume that,
t=sin-1xdt=11-x2dx
Then the integral becomes,
I=∫tdt⇒I=t22+c⇒I=12sin-1x2+cwhere,t=sin-1x
Hence, the correct answer is Option B.
The general solution of a differential equation of the type dxdy+P1x=Q1 is
(a) ye∫P1 dy=∫(Q1e∫P1 dy)dy+C
(b) ye∫P1 dx=∫(Q1e∫P1 dx)dx+C
(c) xe∫P1 dy=∫(Q1e∫P1 dy)dy+C
(d) xe∫P1 dx=∫(Q1e∫P1 dx)dx+C