Question

find integrated.:$\boxed[\mathtt\red{\int\left(a^{x}+a^{-x}\right)^{3} d x}}$​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\mathtt{ \displaystyle\int\left(a^{x}+a^{-x}\right)^{3} d x}$

We know that

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + {3x}^{2}y + {3xy}^{2} + {y}^{3}}}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\int \: \bigg[ {a}^{3x} + 3{a}^{2x} {a}^{ - x} + 3{a}^{x}{a}^{ - 2x} + {a}^{ - 3x}\bigg]dx$

$\rm \:  =  \: \displaystyle\int \: \bigg[ {a}^{3x} + 3{a}^{x} + 3{a}^{ - x}+ {a}^{ - 3x}\bigg]dx$

We know,

$\green{\boxed{\tt{ \displaystyle\int \: {a}^{mx + c}dx \: = \: \frac{{a}^{mx + c}}{m \: loga} \: + d }}}$

$\rm \:  =  \: \dfrac{{a}^{3x}}{3 \: loga} + 3\dfrac{{a}^{x}}{loga} + 3\dfrac{{a}^{ - x}}{ - loga} + \dfrac{{a}^{ - 3x}}{ - 3 \: loga} + c$

$\rm \:  =  \: \dfrac{1}{3 \: loga} \: \bigg[{a}^{3x} + 9{a}^{x} - 9{a}^{ - x} - {a}^{ - 3x}\bigg] + c$

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LEARN MORE

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

Answer 2

Step-by-step explanation:

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

Given integral is

$\rm :\longmapsto\:\mathtt{ \displaystyle\int\left(a^{x}+a^{-x}\right)^{3} d x}$

We know that

$\red{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + {3x}^{2}y + {3xy}^{2} + {y}^{3}}}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\int \: \bigg[ {a}^{3x} + 3{a}^{2x} {a}^{ - x} + 3{a}^{x}{a}^{ - 2x} + {a}^{ - 3x}\bigg]dx$

$\rm \:  =  \: \displaystyle\int \: \bigg[ {a}^{3x} + 3{a}^{x} + 3{a}^{ - x}+ {a}^{ - 3x}\bigg]dx$

We know,

$\blue{\boxed{\tt{ \displaystyle\int \: {a}^{mx + c}dx \: = \: \frac{{a}^{mx + c}}{m \: loga} \: + d }}}$

$\rm \:  =  \: \dfrac{{a}^{3x}}{3 \: loga} + 3\dfrac{{a}^{x}}{loga} + 3\dfrac{{a}^{ - x}}{ - loga} + \dfrac{{a}^{ - 3x}}{ - 3 \: loga} + c$

$\rm \:  =  \: \dfrac{1}{3 \: loga} \: \bigg[{a}^{3x} + 9{a}^{x} - 9{a}^{ - x} - {a}^{ - 3x}\bigg] + c$

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LEARN MORE

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