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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 2

Answers (1)

Answer:

               \frac{1}{14}\sin7x+\frac{1}{2}\sin x+c

Hint:

               2\cos A\cos B=\cos\left ( A+B \right )+ cos \left ( A-B\right )

Given:

               \int \cos3x\cos4xdx

Explanation:

\begin{aligned} & \int \cos 3 x \cos 4 x d x=\int \frac{1}{2}[\cos (3 x+4 x)+\cos (3 x-4 x)] d x \\ =& \frac{1}{2} \int \cos 7 x d x+\frac{1}{2} \int \cos x d x \\ =& \frac{1}{2} \times \frac{\sin 7 x}{7}+\frac{1}{2} \sin x+c \\ =& \frac{1}{14} \sin 7 x+\frac{1}{2} \sin x+c \end{aligned}

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