Calculate $\int_{0}^{\pi} \frac{x}{a-\sin{x}}dx , \quad a>1$

I have trouble calculating this integral.

I tried integration by parts and trigonometric function.

$$\int_{0}^{\pi} \frac{x}{a-\sin{x}}dx , \quad a>1$$

Let the integral be $I$.

A useful trick with this sort of thing is to substitute $y=\pi-x$. Then we have $$I = \int_0^{\pi} \frac{(\pi-y)}{a-\sin{y}} \, dy,$$ since $\sin{(\pi-y)}=\sin{y}$. Hence, adding, we have $$2I = \pi\int_0^{\pi} \frac{dx}{a-\sin{x}}.$$ To finish the computation, you can either use the tangent–half-angle substitution (i.e. $t=\tan{(x/2)}$), or do a residue calculation: if you know one of these techniques, the calculation from here is straightforward.

• All too easy. +1 And happy holidays! - Mark – Mark Viola Dec 24 '15 at 5:27

General Case

In these types of definite integrals, never forget to use this general identity

$$I=\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$

Which can be proved by the substitution $x \to a+b-x$. Then one writes the definite integral as the average of the two expressions above to obtain

$$I=\int_{a}^{b} \frac{1}{2}[f(x)+f(a+b-x)]dx$$

Next, the magic happens since you can find the primitive of $g(x)=\frac{1}{2}[f(x)+f(a+b-x)]$ but not that of $f(x)$ or $f(a+b-x)$. This is usually due to the simpler form of $g(x)$ in comparison with $f(x)$ or $f(a+b-x)$ as some expression has cancelled or disappeared in $g(x)$.

In your example we have

\begin{align} a &= 0 \\ b &= \pi \\ f(x) &= \frac{x}{a-\sin(x)} \\ f(a+b-x) &= \frac{\pi-x}{a-\sin(\pi-x)} = \frac{\pi-x}{a-\sin(x)} \\ g(x) &= \frac{\pi}{2} \frac{1}{a-\sin(x)} \end{align}

Can you see the cancellation that is happening in $g(x)$? Then the integral becomes

$$I=\frac{\pi}{2}\int_{0}^{\pi} \frac{1}{a-\sin(x)} dx$$

• Case $|a| \gt 1$

Then you can find by tangent half angle substitution $u=\tan(\frac{x}{2})$ that

$$F(x)=\int \frac{1}{a-\sin(x)} = \frac{2}{\sqrt{a^2-1}} \arctan\left(\frac{a \tan(\frac{x}{2})-1}{\sqrt{a^2-1}}\right) + C$$

As you can see this formula is valid for $|a| \gt 1$. Hence, the final result will be

$$I=\frac{\pi}{\sqrt{a^2-1}} \left(\arctan\left(\frac{1}{\sqrt{a^2-1}}\right)+\frac{\pi}{2}\right)$$

• Case $|a| \lt 1$

I will leave this case as an exercise for you. The procedure is the same but just the $F(x)$ will be different. However, $F(x)$ is obtained with the same technique for substitution.

• Case $|a|=1$

This is also another case, which should be handled separately. The $F(x)$ in this case is the simplest one and is obtained with the same techniques.

• When you have fraction inside parentheses, using $\left(\right)$ is better. Compare $\left(\arctan\left(\frac{1}{\sqrt{a^2-1}}\right)+\frac{\pi}{2}\right)$ with $(\arctan(\frac{1}{\sqrt{a^2-1}})+\frac{\pi}{2})$ – user2838619 Dec 24 '15 at 16:00
• @user2838619: Thanks for suggested type setting. Will do it. :) – H. R. Dec 24 '15 at 16:01