Question

integrate 2 sin x + cos 3x dx​

Answer 1

Step-by-step explanation:

-2 cosx + 3 sin3x +c

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Answer 2

To find :

$\displaystyle\sf{\int{ \; 2 sin (x)\; .\;cos (3x) \;\;dx =\; ?}}$

AnswEr :

Given expression is :

$\implies\displaystyle \sf{\int{\;2\;sin(x)\;.\;cos(3x)\;\;dx}}$

using trigonometric identity

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

$\implies\displaystyle \sf{\int{sin(x+3x)+sin(x-3x)\;\;dx}}$

$\implies\displaystyle \sf{\int{sin(4x)+sin(-2x)\;\;dx}}$

using trigonometric identity

sin (-A) = - sin A

$\implies\displaystyle \sf{\int{sin(4x)-sin(2x)\;\;dx}}$

$\implies\displaystyle \sf{\int{sin(4x)\;.\;dx}-\int{sin(2x)\;.\;dx}}$

since,

$\displaystyle\sf{\int\;sin(ax+b)=\dfrac{-1}{a}cos(ax+b)+c}$

(where c is an arbitary constant)

therefore,

$\implies\displaystyle \sf{\bigg(\dfrac{-1}{4}cos(4x)\bigg)-\bigg(\dfrac{-1}{2}cos(2x)\bigg)+c}$

$\boxed{\boxed{\implies\displaystyle \sf{\dfrac{-1}{4}cos(4x)+\dfrac{1}{2}cos(2x)+c}}}$

Hence,

• Integral of 2 sin (x) cos (3x) dx is $\bf{\dfrac{-1}{4}cos(4x) +\dfrac{1}{2}cos(2x) + c}.$