Math, asked by Anonymous, 1 day ago


 { \boxed{\bold \red{ \int \cos(3x)dx}}}




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Answers

Answered by sajan6491
15

Let u=3 x.

 \bold \red{\color{red}{\int{\cos{\left(3 x \right)} d x}} = \color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}

{ \bold \red{\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du \: with  \: c=\frac{1}{3} \:  and \:  f{\left(u \right)} = \cos{\left(u \right)}}}

\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}} = \color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}

 \bold \red{ \frac{\color{red}{\int{\cos{\left(u \right)} d u}}}{3} = \frac{\color{red}{\sin{\left(u \right)}}}{3}}

 \bold \red{ \frac{\sin{\left(\color{red}{u} \right)}}{3} = \frac{\sin{\left(\color{red}{\left(3 x\right)} \right)}}{3}}

Therefore,

 \bold \red{\int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}}

 \bold \red{ \int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}+C}

Answered by mathdude500
6

\large\underline{\sf{Solution-}}

Given integral is

\red{\rm :\longmapsto\:\displaystyle\int \:cos(3x) \: dx \: }

To evaluate this integral, we use Method of Substitution

So, Let substitute

\red{\rm :\longmapsto\:3x = y \: }

On differentiating both sides w. r. t. x, we get

\red{\rm :\longmapsto\:3dx = dy \: }

\red{\rm :\longmapsto\:dx = \dfrac{dy}{3} \: }

So, on substituting all these values in above integral, we get

\rm \:  =  \: \displaystyle\int \:cosy \: \dfrac{dy}{3}

\rm \:  =  \: \dfrac{1}{3} \: \displaystyle\int \:cosy \: dy

\rm \:  =  \: \dfrac{1}{3} \: siny \:  +  \: c

\rm \:  =  \: \dfrac{1}{3} \: sin(3x) \:  +  \: c

Hence,

\rm \implies\:\boxed{ \tt{ \: \displaystyle\int \:cos(3x)dx \: =  \: \dfrac{1}{3} \: sin(3x) \:  +  \: c \: }}

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Additional Information :-

 \green{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}}

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