Answer to Question #176980 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

∫ dy/√(25+9y²)

Answer 1

Solution.

$$\int\frac{dy}{\sqrt{25+9y^2}}$$

Make a substitution: $u=\frac{3y}{5},$ then $\frac{du}{dy}=\frac{3}{5},$ since $dy=\frac{5}{3}du.$

So, $\int\frac{dy}{\sqrt{25+9y^2}}=\frac{1}{3}\int \frac{du}{\sqrt{u^2+1}}.$

Make a substitution: $u=\tan{v},$ then $v=\arctan{u},$ since $du=\sec^2{v}dv.$

So,

$\frac{1}{3}\int \frac{du}{\sqrt{u^2+1}}=\int \frac{\sec^2{v}dv}{\sqrt{\tan^2{v}+1}}= \int \sec{v}dv=\newline =\int \frac{\sec{v}(\tan{v}+\sec{v})}{\tan{v}+\sec{v}}dv.$

Make a substitution: $w=\tan{v}+\sec{v},$ then $\frac{dw}{dv}=\sec{v}\tan{v}+\sec^2{v},$ since $dv=\frac{1}{\sec{v}\tan{v}+\sec^2{v}}dw.$

So,

$\int \frac{\sec{v}(\tan{v}+\sec{v})}{\tan{v}+\sec{v}}dv=\int \frac{1}{w}dw=\ln|w|+C= \newline =\ln |\tan{v}+\sec{v}|+C=\ln |\sqrt{u^2+1}+u|+C=\frac{1}{3}\ln |\sqrt{9y^2+25}+3y|+C,$

where $C$ is some constant.

Answer. $\int\frac{dy}{\sqrt{25+9y^2}}=\frac{1}{3}\ln |\sqrt{9y^2+25}+3y|+C.$