# Show that $\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$

Show that $$\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$$

So I have attempted this, but have got a different answer so I have obviously done something wrong but i can't spot it. My workings are as shown. Any help will be great.

I started by writing $$\cos\theta + \sqrt{3}\sin\theta$$ as $$2\cos(\theta - \frac \pi 3)$$. This means that integral becoems, $$\int_0^{\pi/3} \frac{1}{2\cos^2(\theta - \frac\pi3)}d\theta = \int_0^{\pi/3} \frac12\sec^2(\theta - \frac\pi3)d\theta = \left[\frac12 \tan(\theta-\frac\pi3)\right]_0^{\pi/3}$$ $$= \frac12\tan(0) - \frac12\tan(-\frac\pi3) = 0 - \frac12\sqrt3$$ So I get a final answer of $$-\sqrt3/2$$ any help will be great.

• Why don't you use the Weierstrass substitution? Commented May 9, 2019 at 18:32
• @Dr.SonnhardGraubner I think Weierstrass is good but over complicates problems like these, it is best when there is a number added to the trigonometric function in the denominator, as in those problems this method cannot be used Commented May 9, 2019 at 19:04

\begin{align*} \int_0^{\pi/3}\frac{1}{(\cos t +\sqrt{3}\sin t)^2} dt=&\frac 14 \int_0^{\pi/3} \frac{1}{\left(\frac 12 \cos t + \frac{\sqrt{3}}{2} \sin t\right)^2} dt = \frac 14 \int_0^{\pi/3}\frac{1}{\cos^2(t -\frac{\pi}{3})}dt\\ =& \frac 14 \left[\tan(t-\frac{\pi}{3})\right]_0^{\pi/3} =\frac{1}{4}(0-(-\sqrt{3})) = \frac{\sqrt{3}}{4} \end{align*}
When you squared the entire expression, you didn't square the $$2$$, which gives you the missing factor of $$4$$. You also messed up the minus sign on the final step, but your evaluation is correct.$$-\tan\left(-\frac\pi3\right)=\sqrt 3$$
Your mistake is in your first attempts to solve the problem $$(2cos(θ-(\pi/3)))^2=4cos^2(θ-(\pi/3))$$ you wrote 2 instead of 4