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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 4

Answers (1)

Answer:

               \frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}-\frac{\cos (m-n) x}{m-n}\right]+c

Hint:

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

Given:

               \sin mx\cos nxdx,m\neq n

Explanation:

\begin{aligned} &\int \sin m x \cos n x d x=\frac{1}{2} \int[\sin (m+n) x d x+\sin (m-n) x d x] \\ &=\frac{1}{2} \int \sin (m+n) x d x+\frac{1}{2} \int \sin (m-n) x d x \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}\right]-\frac{1}{2}\left[\frac{\cos (m-n) x}{m-n}\right]+c \\ &=\frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}-\frac{\cos (m-n) x}{m-n}\right]+c \end{aligned}

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