How to calculate the integral $ \int_{-2}^{2} e^x\sqrt{4-x^2} \mathrm{d}\, x$?

Question

How to calculate the following integral: $$ I=\int_{-2}^{2} e^x\sqrt{4-x^2} \,\mathrm{d} x = ? $$

I try to use the substitution $x=2\sin\theta$ with $\theta\in[-\pi/2,\pi/2]$, and I get $$ I=4\int_{-\pi/2}^{\pi/2} e^{2\sin\theta} \cos^2\theta\,\mathrm{d}\theta. $$ Is there a explicit form of the integral with the form: $$ \int e^{2\sin\theta} \cos^n\theta\,\mathrm{d}\theta,\qquad n\in\mathbb{Z}. $$

Please show your attempts. Users don't like this website being used as a homework machine. — AryanSonwatikar, May 18 '20 at 06:41
This isn't something that can be found analytically; you have to resort to numerical integration. — Decaf-Math, May 18 '20 at 06:45

score 3 · Answer 1 · answered May 18 '20 at 08:13

You properly wrote $$I=4\int_{-\frac\pi 2}^{\frac\pi 2} e^{\sin(\theta)} \cos^2(\theta)\,d\theta$$ Now, let $t=\theta+\frac \pi 2$ to make $$I=4 \int_0^\pi e^{2 \cos (t)} \sin ^2(t)\,dt=2\int_0^\pi e^{2 \cos (t)} \,dt-2\int_0^\pi e^{2 \cos (t)} \cos(2t)\,dt $$ and now we face modified Bessel functions of the first kind. This makes $$I=2 \pi I_0(2) -2 \pi I_2(2)=2 \pi I_1(2)$$

How to calculate the integral $ \int_{-2}^{2} e^x\sqrt{4-x^2} \mathrm{d}\, x$?

1 Answers1