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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 6

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 6

Answers (1)

Answer:

              \frac{x}{4}+\frac{1}{16}\sin4x-\frac{1}{24}\sin6x-\frac{1}{8}\sin2x+c

Hint:

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

Given:

              \sin x\cos2x\sin3xdx

Explanation:

\sin x\cos 2x\sin3xdx=\frac{1}{2}[2 \sin x \cos 2x]\sin3x

                                             =\frac{1}{2}(\sin^{2} 3x-\sin x \sin3x) \quad \quad \quad \quad [2 \sin A \cos B = \sin(A + B) + \sin(A - B)]

                                               \begin{aligned} &=\frac{1}{2} \int \sin ^{2} 3 x d x-\frac{1}{2} \int \sin x \sin 3 x d x \\ &=\frac{1}{2} \int \frac{1-\cos 6 x}{2} d x-\frac{1}{4} \int[-\cos 4 x+\cos 2 x] d x \\ &=\frac{1}{4} \int d x-\frac{1}{4} \int \cos 6 x d x+\frac{1}{4} \int \cos 4 x d x-\frac{1}{4} \int \cos 2 x d x \\ &=\frac{x}{4}+\frac{1}{16} \sin 4 x-\frac{1}{24} \sin 6 x-\frac{1}{8} \sin 2 x+c \end{aligned}

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