Evaluate $\int \frac{1}{(a^{2} + y^{2})^{3/2}}dy$

Question

Evaluate $$\int \frac{1}{(a^{2} + y^{2})^{3/2}}dy$$

where $a$ is a constant.

Any hints would do.

Answer 1

Using hyperbolic trig we get $1+\sinh^2\theta=\cosh^2\theta$ and using that we can make the substitution $x=a\sinh\theta$ and $\text{d}x=a\cosh\theta\text{d}\theta$ which changed our integral into $\int\frac{\text{sech}^2\theta}{a^2}\text{d}\theta=\frac{\tanh\theta}{a^2}=\boxed{\frac{x}{a^2\sqrt{x^2+a^2}}+C}$

Answer 2

Hint : Substitute $u=\dfrac1{y^2}$.

Answer 3

Hint

Let $y=a \tan (t)$ , $dy=\frac{a \mathrm d t}{\cos ^2 (t)}$, so your integral becomes: \begin{align} \int \frac{a}{\cos^2(t)}\frac{\mathrm d t}{(a^2+a^2\tan^2(t))^{3/2}} \end{align} Note that $1+\tan^2(t)=\frac{1}{\cos^2(t)}$

Answer 4

$$\int f(\phi(t))\phi' (t)~dt = F(\phi(t)) + C \tag{1}$$ where $F$ is an antiderivative of $f$ and $C$ is the constant of integration.

Set $f(y) = \dfrac{1}{(a^2 + y^2)^{3/2}}$ and $\phi (t) = y = a\tan t$. So, you get $$f(\phi(t))\phi'(t) = \dfrac{1}{a^2\sec t} \tag{2}$$

Substitute the value of $(2)$ in $(1)$ and solve the integral. You will get the integral in terms of $t$. Use $y = a\tan t$ to get $t = \arctan\left(\dfrac{y}{a}\right)$ and substitute it in the answer.

Also note that $\sin(\arctan(z)) = \dfrac{z}{\sqrt{z^2 + 1}}$