Evaluate the following $\int\frac{dx}{\sqrt {16-9x^2}}$

Question

$\begin{array}{1 1}(A)\;\large\frac{1}{3} \cos ^{-1} \bigg(\frac{3x}{4}\bigg)+c \\(B)\;\large\frac{1}{3} \sin ^{-1} \bigg(\frac{4x}{3}\bigg)+c \\ (C)\;\large\frac{1}{3} \cos ^{-1} \bigg(\frac{4x}{3}\bigg)+c \\ (D)\;\large\frac{1}{3} \sin ^{-1} \bigg(\frac{3x}{4}\bigg)+c\end{array} $

Answer 1

Toolbox:

• $\large\int\frac{dx}{\sqrt {a^2-x^2}}=\sin ^{-1} (\frac{x}{a})+c$

$\large\int\frac{dx}{\sqrt {16-9x^2}}$

$\large\int\frac{dx}{\sqrt {16-9x^2}}=\sqrt {9(\frac{16}{9}-x^2)}$

$=3 \sqrt {(4/3)^2-x^2}$

This is of the form $\large\int\frac{dx}{\sqrt {a^2-x^2}}=\sin ^{-1} (\frac{x}{a})+c$

Here $a=\frac{4}{3}\; and\; x=x$

Hence $\large\frac{1}{3} \int \frac{dx}{(\frac{4}{3})^2-x^2}$ on integrating gives

$=\large\frac{1}{3} \int \frac{dx}{\sqrt {(\frac{4}{3})^2-x^2}}$

$=\large\frac{1}{3} \sin ^{-1} \bigg(\frac{x}{\frac{4}{3}}\bigg)+c$

$=\large\frac{1}{3} \sin ^{-1} \bigg(\frac{3x}{4}\bigg)+c$