∫ 4dx/√1-16x^2 =

Question

\(\int \frac{4dx}{\sqrt{1-16x^2}}=\)

(A) cos^-14x + K

(B) 1/4 cos^-14x + K

(C) sin^-14x + K

(D) 4sin^-14x + K

Answer 1

Correct option is (C) sin^-14x + K 

\(\int \frac{4dx}{\sqrt{1-16x^2}}dx= \int \frac{dx}{\sqrt{(\frac 14)^2-x^2}}\)

\(= sin^{-1} \left(\frac x{\frac 14}\right) +K\)

\(= sin^{-1} (4x) + K\)