What is $\int_{-2}^2\frac{1.4^x\sqrt{4-x^2}}{1.8}dx$ Evaluate the integral:
$$\int_{-2}^2\frac{1.4^x\sqrt{4-x^2}}{1.8}dx$$
I have tried integration by parts but end up in a loop of integration. I have also entered the integral into WolframAlpha and recieved an answer of approximately $3.692$.
Here are my questions:

*

*Is manual integration possible?

*If so, how would I go about solving the integral?

Thanks in advance
 A: You can obtain a solution using special functions.
From the Digital Library of Mathematical Functions, we know that
$$
\frac{\pi}{\color{purple}{z}}I_1(\color{purple}{z}) = \int_{-1}^{1}e^{\color{purple}{z}t} \sqrt{1-t^2}\, \mathrm{d}t \tag{1}
$$
So for $a,b >0$ we get
$\require{\cancel}$
\begin{align}
\int_{-a}^{a}b^x \sqrt{a^2-x^2}\, \mathrm{d} x & \overset{t= x/a}{=} \int_{-1}^{1}b^{at} a \sqrt{1-t^2}\, a\mathrm{d}t\\
& = a^2\int_{-1}^{1} e^{\left[\color{purple}{a\ln(b)}\right] t} \sqrt{1-t^2}\, \mathrm{d}t\\
& \overset{(1)}{=} a^{\cancel{2}} \frac{\pi}{\cancel{\color{purple}{a}}\color{purple}{\ln(b)}} I_1(\color{purple}{a\ln(b)})\\
& = \pi a\frac{I_1\left(a\ln\left(b\right)\right)}{\ln(b)}
\end{align}
And applied to your case we get
$$
\int_{-2}^{2} \frac{\left(\frac{7}{5}\right)^x\sqrt{4-x^2}}{\left( \frac{9}{5}\right)}\, \mathrm{d} x = \frac{10\pi}{9}\frac{I_1\left(2\ln\left(\frac{7}{5}\right)\right)}{\ln\left(\frac{7}{5}\right)} \approx  3.692017232642\dots
$$
where you can verify the numerical approximation at the end using WA.
