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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 1

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.7 Question 1

Answers (1)

Answer:

            -\frac{1}{22}\cos 11x+\frac{1}{6}\cos3x+c

Hint:

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

Given:

                \int \sin4x\cos7x dx

Explanation:
I=\frac{1}{2}\int2 \sin4x\cos7xdx=\frac{1}{2}\int\[sin\left ( 4x+7x \right )+\sin\left ( 4x-7x \right )]dx\\ =\frac{1}{2}\int \sin11x dx-\frac{1}{2}\int \sin 3x dx \\ =\frac{1}{2}\left [ \frac{\cos11x}{-11} \right ]-\frac{1}{2}\left [ \frac{\cos3x}{-3} \right ]+c\\ =\frac{-1}{22}\cos11x+\frac{1}{6}\cos3x+c

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