MHT CET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{6}}\left(2+3 x^2\right) \cos 3 x d x=\)
- A \(\frac{2}{9}+\frac{\pi^2}{36}\)
- B \(\frac{4}{9}+\frac{\pi^2}{36}\)
- C \(\frac{2}{9}-\frac{\pi^2}{36}\)
- D \(\frac{4}{9}-\frac{\pi^2}{36}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{9}+\frac{\pi^2}{36}\)
Step-by-step Solution
Detailed explanation
\(\int_0^{\frac{\pi}{6}}\left(2+3 x^2\right) \cos 3 x d x = \left[\left(x^2+\frac{4}{9}\right) \sin 3x + \frac{2x}{3} \cos 3x\right]_0^{\frac{\pi}{6}}\) \(= \left(\left(\frac{\pi}{6}\right)^2+\frac{4}{9}\right) \sin\left(\frac{3\pi}{6}\right) + \frac{2(\frac{\pi}{6})}{3} \cos\left(\frac{3\pi}{6}\right) - \left(\left(0\right)^2+\frac{4}{9}\right) \sin(0) - \frac{2(0)}{3} \cos(0)\)
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