Computation of $\int{\sqrt{c^2+b^2t^2}\,\mathrm dt}$.

Question

How to solve $$\int{\sqrt{c^2+b^2t^2}\,\mathrm dt}$$

I tried substituting $t$ with $\sin{x}$ but it doesn't work, since $b^2$ creates a problem.

Answer 1

Hint: Right approach, wrong substitution choice.

Try putting $$bt = c\tan x \iff t = \frac cb\tan x \implies dt = \frac cb \sec^2 x\,dx$$

Answer 2

Another hint :

Try using hyperbolic functions. Let $t=\dfrac cb\sinh x\;\Rightarrow\;dt=\dfrac cb\cosh x\ dx$ and use identity $\cosh^2x=1+\sinh^2x$ and $\cosh^2x=\dfrac{1+\cosh2x}{2}$.

Answer 3

$$\int \sqrt{c^2 + b^2 t^2} dt = \int \sqrt{b^2\left(\frac{c^2}{b^2}+ t^2\right)} dt = |b| \int \sqrt{\left(\frac{c}{b}\right)^2+ t^2} dt$$

Let $a = \frac{c}{b}$, and you get:

$$=|b| \int \sqrt{t^2 + a^2} dt$$

Now you can use techniques from this previous question:

Integral $\int \sqrt{x^{2}+a^{2}}$