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Solve the following.
Evaluate the following integrals: Evaluate: ∫ tan x dx. Hence, evaluate: ∫ (dx)/(cotx/2 cotx/3…
Evaluate the following integrals:
(a)[2]
Evaluate: $\int \tan x dx$.
(b)[3]
Hence, evaluate: $\int \frac{dx}{\cot\frac{x}{2} \cot\frac{x}{3} \cot\frac{x}{6}}$.
Answer
Answer (a)
Official answer key$\int \tan x\,dx = \int \frac{\sin x}{\cos x}dx = -\int \frac{d(\cos x)}{\cos x} = -\ln|\cos x| + c$.
Final answer: $-\ln|\cos x| + c$
Answer (b)
Official answer key$\frac{1}{\cot\frac{x}{2}\cot\frac{x}{3}\cot\frac{x}{6}} = \tan\frac{x}{2}\tan\frac{x}{3}\tan\frac{x}{6}$.
Since $\frac{x}{2} = \frac{x}{3} + \frac{x}{6}$: $\tan\frac{x}{2} = \frac{\tan\frac{x}{3} + \tan\frac{x}{6}}{1 - \tan\frac{x}{3}\tan\frac{x}{6}}$, so $\tan\frac{x}{2}\tan\frac{x}{3}\tan\frac{x}{6} = \tan\frac{x}{2} - \tan\frac{x}{3} - \tan\frac{x}{6}$.
Using part (a): $\int \tan\frac{x}{k}dx = -k\ln\left|\cos\frac{x}{k}\right|$, so the integral is $-2\ln\left|\cos\frac{x}{2}\right| + 3\ln\left|\cos\frac{x}{3}\right| + 6\ln\left|\cos\frac{x}{6}\right| + c$.