∫sec2sin-1x1-x2dx=
sintan-1x+c
tansec-1x+c
tansin-1x+c
-tancos-1x+c
Explanation for correct option:
Evaluate ∫sec2sin-1x1-x2dx
Consider the given equation as
I=∫sec2sin-1x1-x2dx
Let us assume that,
sin-1x=t⇒11-x2dx=dt
Substitute the above values in the integral becomes
I=∫sec2tdtI=tant+cI=tansin-1x+cwhere,t=sin-1x
Hence, the correct answer is option C.
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