Differential equations: tricky integral involving exponential and trigonometric functions

Question

While doing differential equations homework the integral $$\int \exp\left(\sec^2x\right)\, 2x\cos(x) \,dx$$ popped up and I've been struggling with it for a while.

Any help with integral would be much appreciated.

Answer 1

Usually the exponent appears in derivatives of exponential function, so we take derivative of $\exp(\sec^2 x)$ and compare it with integrant to estimate the integral.

$$ e^{\sec^2 x}=e^ \frac{1}{\cos^2 x}$$

$$\left(e^ \frac{1}{\cos^2 x} \right)'=-\frac{2\sin x \cos x}{\cos^4 x}. e^ \frac{1}{\cos^2 x}$$

$$I=\int e^ \frac{1}{\cos^2 x}2 x \cos x dx =\int e^ \frac{1}{\cos^2 x}\frac{-2 \sin x \cos x}{\cos^4 x} dx \cdot \left(\frac{-x \cos^4 x}{\sin x}\right) $$

Integrating by parts we have:

$$I=\left(\frac{-x \cos^4 x}{\sin x} \right) \cdot \left(e^ \frac{1}{\cos^2 x} \right)- \int \left(\frac{-x \cos^4 x}{\sin x} \right)'.e^ \frac{1}{\cos^2 x} dx$$

you can continue integrating but I do not think it can have a closed form.