Question

Integration of Sin (s+x) sin(x+t)dx

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int \: sin(s + x) \: sin(x + t) \: dx$

can be rewritten as

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int \: 2 \: sin(s + x) \: sin(x + t) \: dx$

We know,

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

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int \: [cos(s + x - x - t) - cos(s + x + x + t)] \: x$

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

We know,

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

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int \: k \: dx \: = \: kx \: + \: c \: }}}$

So, using these, we get

$\rm \:  =  \: \dfrac{1}{2}\bigg[x \: cos(s - t) - \dfrac{sin(2x + s + t)}{2} \bigg] + c$

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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

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int \: sin(s + x) \: sin(x + t) \: dx$

can be rewritten as

$\rm \:  =  \: \dfrac{1}{2} \displaystyle\int \: 2 \: sin(s + x) \: sin(x + t) \: dx$

We know,

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

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int \: [cos(s + x - x - t) - cos(s + x + x + t)] \: x$

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

We know,

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

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int \: k \: dx \: = \: kx \: + \: c \: }}}$

So, using these, we get

$\rm \:  =  \: \dfrac{1}{2}\bigg[x \: cos(s - t) - \dfrac{sin(2x + s + t)}{2} \bigg] + c$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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