# Integration by Euler's formula

How do you integrate the following by using Euler's formula, without using integration by parts? $$I=\displaystyle\int \dfrac{3+4\cos {\theta}}{(3\cos {\theta}+4)^2}d\theta$$

I did integrate it by parts, by writing the $$3$$ in the numerator as $$3\sin^2 {\theta}+3\cos^2{\theta}$$, and then splitting the numerator.

But can it be solved by using complex numbers and the Euler's formula?

Hint

When you have an expression with a squared denominator, you could think that the solution is of the form $$I=\displaystyle\int \dfrac{3+4\cos {\theta}}{(3\cos {\theta}+4)^2}~d\theta=\frac{a+b\sin \theta+c\cos \theta}{3\cos {\theta}+4}$$ Differentiate the rhs and identify terms. You will get very simple results.

• is this applicable every time when there is a squared denominator over a linear? – pkwssis Jul 16 '14 at 8:29
• Think about the derivative of $\frac{u(x)}{v(x)}=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}$. Cheers :) – Claude Leibovici Jul 16 '14 at 16:11

By remembering that $e^{i\theta}=\cos(\theta) + i\sin(\theta)$ it is then easy to see that

$$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$$

and

$$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

$$\frac{3+2(e^{i\theta} + e^{-i\theta})}{(4+\frac{3}{2}(e^{i\theta} + e^{-i\theta}))^2}$$
Now you need to expand that expression, and apply the rules you learned from your first course in integral calculus to then solve in terms of $e^{i\theta}$. You then need to play with adding fun values of zero and multiplying by fun values of 1 to repackage into sin and cosine, or, simply just forward apply Euler's magnificent formula to translate back.
• After simplifying and substituting $e^{i\theta}=u$, then what? – pkwssis Jul 16 '14 at 4:55
• Don't substitute $u = e^{i\theta}$, substitute, instead, $u = i\theta$ so that you have familiar expressions of $e^u$ and $e^{-u}$ – Timothy Shoaf Jul 16 '14 at 6:13
Hint: $$\frac{d}{dx}\frac{\sin x}{3\cos x+4}=\frac{3+4\cos x}{(3\cos x+4)^2}.$$