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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2010 · Mathematics · Definite Integration

(∫4 (x+(π)/(6)) 2 x · ((5 π)/(6)+ x) d x)

(\int4 \cos \l(x+π6)cos⁡\l\left(x+\frac{\pi}{6}\right) \cos2 x \cdot \cos \left(\frac{5 \pi}{6}+$x\right) d x)

Options

  1. A(-\left(x+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c)
  2. B(-\left(x+\frac{\sin 4 x}{4}-\frac{\sin 2 x}{2}\right)+c)
  3. C(-\left(x-\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c)
  4. D(-\left(x-\frac{\sin 4 x}{4}+\frac{\cos 2 x}{2}\right)+c)

Correct answer

A. (-\left(x+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c)

Step-by-step solution

$\begin{aligned}& \int4 \cos \left( x+\frac{\pi}{6}\right) \cos2 x \cdot \cos \left(\frac{5 \pi}{6}+ x\right) d x \\ & =2 \int\left(\cos (2 x+\pi) \cos \frac{2 \pi}{3}\right) \cos2 x d x \\ & =2 \int\left(-\cos2 x-\frac{1}{2}\right) \cos2 x d x \\ & =\int\left(-2 \cos ^2 2 x-\cos 2 x\right) d x \\ & =-\int(1+\cos 4 x+\cos 2 x) d x \\ & =-x-\frac{\sin 4 x}{x}-\frac{\sin 2 x}{2}+ c \end{aligned}$)
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