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Find an antiderivative (or integral) of the following function by the method of inspection. sin 2x – 4 e3x

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Question

Find an antiderivative (or integral) of the following function by the method of inspection.

sin 2x – 4 e3x

Sum
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Solution

we know that,

`d/dx` cos 2x = - 2 sin 2x

or sin 2x = `d/dx (- 1/2  "cos 2x")`

and `d/dx  e^(3x) = 3e^(3x)`

or `e^(3x) = d/dx (1/3  e^(3x))`

Hence,  sin 2x - 4e3x 

`= d/dx (- 1/2  cos 2x) - 4  d/dx (1/3 e^(3x))`

 `= d/dx (- 1/2  cos 2x - 4/3 e^(3x))`

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Chapter 7: Integrals - Exercise 7.1 [Page 299]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.1 | Q 5 | Page 299

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