1.

Integrate 2 sin x + cos 3x dx​

Answer»

To FIND :

\displaystyle\sf{\int{ \; 2 <klux>SIN</klux> (x)\; .\;<klux>COS</klux> (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,


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