# Evaluating the integral : $\int_{1}^{2}\frac{x+\tan x}{x+\sin x}dx$

$Q.$ Evaluate the following integral :

$\int_{1}^{2}\frac{x+\tan x}{x+\sin x}dx$. Numerically I found that the answer is roughly $1.000006$ but I am unable to compute using the analytic methods.

I tried first computing by splitting:

$\int_{1}^{2}\frac{x}{x+\sin x}dx+\int_{1}^{2}\frac{\tan x}{x+\sin x}dx$

and then applying by-parts to each of them, but that results in a very difficult task.

• Presumably you mean Cauchy Principal Value, since the integral diverges. – André Nicolas Aug 23 '14 at 6:33

Second, there is a problem at $x=\frac{\pi}{2}$ because of the tangent.
So, as suggested by André Nicolas, consider $$\int_{1}^{2}\frac{x+\tan x}{x+\sin x}dx=\int_{1}^{\frac{\pi}{2}-\epsilon}\frac{x+\tan x}{x+\sin x}dx+\int_{\frac{\pi}{2}+\epsilon}^{2}\frac{x+\tan x}{x+\sin x}dx$$ and look at the limits when $\epsilon$ goes to $0$.
$\quad$ By changing the limits of integration to $\dfrac\pi4$ and $\dfrac{3\pi}4$, the definite integral becomes $\dfrac{2\pi}{2+\pi}$.
$\quad$ Otherwise, given the fact that the integrand does not possess a primitive expressible in terms of elementary functions, coupled with the fact that $1$ and $2$ are “meaningless” arguments for the sine and tangent function, the only approaches for the unmodified definite integral are numerical ones.