Question

Solve the integral of √(49-x²)÷x from ln1 to ln2, using trigonometric substitution.

Answer 1

Let $x=7\sin{u},dx=\cos udu$

$I=\int \frac{\sqrt{49-x^2}}{x}dx=7\int \frac{cos^2u}{sinu}du=\int \frac{sinucos^2u}{1-cos^2u}du$

Let $v=cosu,dv=-sinudu$

$I=-7\int \frac{v^2}{1-v^2}dv=-7\int [-1+\frac{1}{2(v+1)}-\frac{1}{2(v-1)}]dv=$

$=7v-\frac{7}{2}ln|\frac{v+1}{v-1}|+C=$

$=\sqrt{49-x^2}+\frac{7}{2}ln|\frac{\sqrt{49-x^2}-7}{\sqrt{49-x^2}+7}|+C$

Thus, $\int_{ln1}^{ln2}\frac{\sqrt{49-x^2}}{x}dx=(\sqrt{49-x^2}+\frac{7}{2}ln|\frac{\sqrt{49-x^2}-7}{\sqrt{49-x^2}+7}|)|_{x=ln1}^{x=ln2}=\infin$

The integral is divergent.