Question

The integral
\(\int_{\pi/6}^{\pi/3} \tan^3 x \cdot \sin^2 3x\left(2\sec^2 x \cdot \sin^2 3x + 3\tan x \cdot \sin 6x\right)dx\)
is equal to:

• A. \(-\frac19\)
• B. \(-\frac{1}{18}\)
• C. \(\frac{7}{18}\)
• D. \(\frac92\)

Hint:

Let \(h = \tan^2 x\,\sin^2 3x\). The bracket times \(\tan x\) is exactly \(h'\), so the integrand is \(h\,h'\).

Answer 1

The Correct Option is B

Let \(h(x) = \tan^2 x \cdot \sin^2 3x\).
\(h'(x) = 2\tan x\sec^2 x \sin^2 3x + \tan^2 x\cdot 2\sin 3x \cdot 3\cos 3x = \tan x\left(2\sec^2 x\sin^2 3x + 3\tan x \sin 6x\right)\).
The integrand is \(\tan^2 x \sin^2 3x \cdot \tan x(\ldots) = h \cdot h'\).
\(I = \left[\frac{h^2}{2}\right]_{\pi/6}^{\pi/3}\).
\(h(\pi/3) = 3\sin^2\pi = 0\). \(h(\pi/6) = \frac13\sin^2\frac{\pi}{2} = \frac13\).
\(I = 0 - \frac12\cdot\frac19 = -\frac{1}{18}\).
Answer: (2) \(-\frac{1}{18}\)