Question

The value of the integral $\int \frac{\sqrt{c^2-x^2}}{x}$ using Euler's substitution is?

• A. $-c+\sqrt{c^2-x^2}-clog\frac{c+\sqrt{c^2-x^2}}{x}+C_1$
• B. $-c+\sqrt{c^2-x^2}-clog\frac{c-\sqrt{c^2-x^2}}{x}+C_1$
• C. $-c-\sqrt{c^2+x^2}-clog\frac{c+\sqrt{c^2-x^2}}{x}+C_1$
• D. $-c+\sqrt{c^2-x^2}-clog\frac{c-\sqrt{c^2+x^2}}{x}+C_1$

Answer 1

The correct option is A $-c+\sqrt{c^2-x^2}-clog\frac{c+\sqrt{c^2-x^2}}{x}+C_1$

We have to find the value of the integral $\int \frac{\sqrt{c^2-x^2}}{x}dx$ using Euler's substitution.
Consider $\int \frac{\sqrt{c^2-x^2}}{x}dx$
We can use second Euler substitution: $\sqrt{c^2-x^2}=xt-c$ $...\left(1\right)$
$\Rightarrow c^2-x^2=(xt-c{)}^2$
$\Rightarrow c^2-x^2=x^2{t}^2-2xct+c^2$
$\Rightarrow x=\frac{2ct}{t^2+1}$ $...\left(2\right)$
Differentiating both sides we get
$dx=\frac{2c(1+t^2)}{(1+t^2{)}^2}dt$
Also $\sqrt{c^2-x^2}=xt-c=\frac{2ct}{t^2+1}t-c=\frac{c(t^2-1)}{t^2+1}$
$\therefore \int \frac{\sqrt{c^2-x^2}}{x}dx=-c\int \frac{(1-t^2{)}^2}{t(1+t^2{)}^2}dt$
$=c\int (\frac{4t}{(1+t^2{)}^2}-\frac{1}{t})dt$
$\therefore \int \frac{\sqrt{c^2-x^2}}{x}dx=-\frac{2c}{1+t^2}-clogt+C_1$ $...\left(3\right)$
By $\left(2\right)$ we have $\frac{x}{t}=\frac{2c}{t^2+1}$ and
By $\left(1\right)$ we have $t=\frac{c+\sqrt{c^2-x^2}}{x}$
Hence $\left(3\right)$ becomes
$\therefore \int \frac{\sqrt{c^2-x^2}}{x}dx=-\frac{x^2}{c+\sqrt{c^2-x^2}}-clog\frac{c+\sqrt{c^2-x^2}}{x}+C_1$
$=-c+\sqrt{c^2-x^2}-clog\frac{c+\sqrt{c^2-x^2}}{x}+C_1$