We want to find the integrals ∫1√a2−x2 and ∫1|x|√x2−1.
We will find these integrals by method of inspection, trying to guess the integral, like we did for integral of sinx, cosx and many other functions. Since we are given integrals on the RHS, to solve this question, we simply differentiate the functions on the RHS and check if we get the same function in the integral.
We know that derivative of sin−1(x) is 1√1−x2.
But we want to find integral of 1√a2−x2
For this we will take ‘a’ common from the denominator. So we get 1|a|√1−(xa)2. This may be related to the derivative of sin−1(xa).
To check that, let’s find the derivative of sin−1(xa). We get 1a√1−(xa)2, which is same as the integral we want to find.
⇒∫1√a2−x2dx=sin−1(xa)⇒First statement is not correct.2.∫1|x|√x2−1 is the derivative of sec−1(x)⇒∫dx|x|√x2−1=sec−1(x)
So the second statement is correct We can also find these integrals by substituting x = a sin t and x = sec t