Answer: \(\frac{1+\cot^2\theta}{\cot^2\theta}=\sec^2\theta\), \(\frac{\mathrm d}{\mathrm d\theta}\tan\theta=\sec^2\theta\), integral \(=\sqrt3-\frac12\).
Using \(1+\cot^2\theta=\csc^2\theta\),
\(\frac{1+\cot^2\theta}{\cot^2\theta}=\frac{\csc^2\theta}{\cot^2\theta}\).
Now \(\csc^2\theta=\frac{1}{\sin^2\theta}\) and \(\cot^2\theta=\frac{\cos^2\theta}{\sin^2\theta}\).
So
\(\frac{\csc^2\theta}{\cot^2\theta}=\frac{1}{\cos^2\theta}\).
Therefore
\(\frac{1+\cot^2\theta}{\cot^2\theta}=\sec^2\theta\).
Also,
\(\frac{\mathrm d}{\mathrm d\theta}\tan\theta=\sec^2\theta\).
Now consider the integral:
\(\displaystyle \int_0^{\pi/3}\left(\frac{1+\cot^2\theta}{\cot^2\theta}-\sin\theta\right)\,d\theta\)
Using \(\frac{1+\cot^2\theta}{\cot^2\theta}=\sec^2\theta\), this becomes
\(\displaystyle \int_0^{\pi/3}(\sec^2\theta-\sin\theta)\,d\theta\).
An antiderivative is
\(\tan\theta+\cos\theta\).
So
\(\displaystyle \int_0^{\pi/3}(\sec^2\theta-\sin\theta)\,d\theta\)
\(\displaystyle =\left[\tan\theta+\cos\theta\right]_0^{\pi/3}\).
Evaluate the upper limit:
\(\tan\frac{\pi}{3}+\cos\frac{\pi}{3}=\sqrt3+\frac12\).
Evaluate the lower limit:
\(\tan0+\cos0=0+1=1\).
Therefore,
\(\displaystyle \left(\sqrt3+\frac12\right)-1\)
\(\displaystyle =\sqrt3-\frac12\).
Hence the value of the integral is
\(\displaystyle \sqrt3-\frac12\).
