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Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 19 Maths Textbook Solution.

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Answer: \frac{5}{12}

Hint: Use indefinite formula then put the limit to solve this integral

Given: \int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx

Solution:\int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx=\frac{2}{2}\int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx

\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}} 2 \cos x \cos 2 x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}[\cos (x+2 x)+\cos (x-2 x)] d x[2 \cos C \cos D=\cos (C+D)+\cos (C-D)] \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}(\cos 3 x+\cos (-x)) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}(\cos 3 x+\cos x) d x \end{aligned}\left [ \cos \left ( -\theta \right )=\cos \theta \right ]

\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \cos 3 x d x+\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \cos x d x \\ &=\frac{1}{2}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{6}} \\ &=\frac{1}{6}[\sin 3 x]_{0}^{\frac{\pi}{6}}+\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{6}} \\ &=\frac{1}{6}\left[\sin 3 \times \frac{\pi}{6}-\sin 3 \times 0\right]+\frac{1}{2}\left[\sin \frac{\pi}{6}-\sin 0\right] \end{aligned}

\begin{aligned} &=\frac{1}{6}\left[\sin \frac{\pi}{2}-\sin 0\right]+\frac{1}{2}\left[\sin \frac{\pi}{6}\right] \\ &=\frac{1}{6} \cdot 1+\frac{1}{2} \cdot \frac{1}{2} \\ &=\frac{1}{6}+\frac{1}{4} \\ &=\frac{4+6}{24}=\frac{10}{24}=\frac{5}{12} \end{aligned}

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