How to integrate $\int{e^{\tan^2{x}}\sin(4x)}dx$

Question

The question says it all. How do I even go about integrating this integral. I'll appreciate some help. Thanks in advance. $$\int{e^{\tan^2{x}}\sin(4x)}dx$$

Answer 1

Let's assume the answer is given by a function of the form: $$I(x) = \exp(\tan^2 x) f(x)$$ Then: $$\frac{dI}{dx} = \exp(\tan^2 x) \left(2 f(x) \sec^2 x \tan x + f'(x)\right)$$ Now we have to solve the ODE: $$2 f(x) \sec^2 x \tan x + f'(x) = \sin(4x)$$ You could use a Fourier Series method to find the solution: $$f(x)=-(4 \cos(2 x)+\cos(4 x)+3)/4=-2\cos^4x$$

Answer 2

Here's another method. First of all, multiply the numerator and denominator by the derivative of $\tan^2{x}$ which comes out something like this $$\int{\frac{\sin{4x}\cdot e^{\tan^2{x}}\cdot 2\tan{x}\sec^2{x}}{2\tan{x}\sec^2{x}}dx}$$ Then, simplifying the $\sin{4x}$ in the numerator and $2\tan{x}\sec^2{x}$ in the denominator using trigonometric identities, we get something like this $$\int{2\cos^4{x}\cdot (2\cos^2{x}-1)\cot e^{\tan^2{x}}\cdot 2\tan{x}\sec^2{x}}dx$$ Then, substituting $t$ for $\tan^2{x}$ and therefore $\frac{1}{1+t}$ for $\cos^2{x}$, you get $$-2\int{\frac{e^t}{(1+t)^2}-\frac{2e^t}{(1+t)^3}}dt$$ which is easily evaluated and comes out to be $$\frac{-2e^t}{(1+t)^2}+C$$ Substituting back $\tan^2{x}$ and $\cos^2{x}$, we get $$-2\cos^4{x}\cdot e^{\tan^2{x}}+C$$