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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 5

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 5

Answers (1)

Answer:

               \frac{-1}{32} \cos 8 x+\frac{1}{48} \cos 12 x-\frac{1}{6} \cos 4 x+c

Hint:     

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

Given:

             \int \sin 2 x \sin 4 x \sin 6 x d x

Explanation:

\begin{gathered} \int \sin 2 x \sin 4 x \sin 6 x d x=\frac{1}{2} \int \sin 2 x(2 \sin 4 x \sin 6 x d x) \\ =\frac{1}{2} \int \sin 2 x[\cos 10 x-\cos 2 x] d x \\ =\frac{1}{2} \int \sin 2 x \cos 10 x d x-\frac{1}{2} \int \sin 2 x \cos 2 x d x \end{gathered}

\begin{aligned} &=\frac{1}{4} \int[-\sin 12 x+\sin 8 x] d x+\frac{1}{4} \int[\sin 4 x-\sin 0 x] d x \\ &=\frac{1}{4}\left[\frac{\cos 12 x}{12}-\frac{\cos 8 x}{8}\right]-\frac{1}{4} \frac{\cos 4 x}{4}+c \\ &=\frac{-1}{32} \cos 8 x+\frac{1}{48} \cos 12 x-\frac{1}{6} \cos 4 x+c \end{aligned}

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