Question

$\red{ \boxed{ \tt \red{ find \: integrals : \purple{\int cos(3x) sin(x) } } }}$
dont Spam :--
Anyone who gives an irrelevant answer will be reported at that moment.
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm cos3x \: sinx \: dx$

On multiply and divide by 2, we get

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm 2 \: cos3x \: sinx \: dx$

We know

$\boxed{\tt{ 2sinx\: cosy\: = \: sin(x + y) + sin(x - y) \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin(x + 3x) + sin(x - 3x)] \: dx$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin4x + sin( - 2x)] \: dx$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin6x - sin 2x ] \: dx$

We know,

$\boxed{\tt{ \displaystyle\int\rm sin(ax + b) \: dx \: = - \frac{cos(ax + b)}{a} + c}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\bigg[ - \dfrac{cos6x}{6} + \dfrac{cos2x}{2} \bigg] + c$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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

Answer 2

Answer:

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm cos3x \: sinx \: dx$

On multiply and divide by 2, we get

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int\rm 2 \: cos3x \: sinx \: dx$

We know

$\boxed{\tt{ 2sinx\: cosy\: = \: sin(x + y) + sin(x - y) \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin(x + 3x) + sin(x - 3x)] \: dx$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin4x + sin( - 2x)] \: dx$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm [sin6x - sin 2x ] \: dx$

We know,

$\boxed{\tt{ \displaystyle\int\rm sin(ax + b) \: dx \: = - \frac{cos(ax + b)}{a} + c}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\bigg[ - \dfrac{cos6x}{6} + \dfrac{cos2x}{2} \bigg] + c$

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$