Mathematics

# Solve:$\int {{{\cos }^3}x{e^{\log \left( {\sin x} \right)}}dx}$

##### SOLUTION

$I=\displaystyle \int \cos^3x\ e^{\log (\sin c)}dx$
put $\sin x=t$
$\cos x \ dx=dt$
$\begin{matrix} \Rightarrow I=\int (1-t^{ 2 })e^{ \log (t) }dt= & \int e^{ \log (t) }dt- & \int t^{ 2 }e^{ \log (t) }dt \\ & \downarrow & \downarrow \\ & \left( A \right) & \left( B \right) \end{matrix}$
$(B)$
$k=\displaystyle \int t^2\ e^{\log (t)}dt$
$k=e^{\log (t)}\dfrac {t^3}{3} \displaystyle \int \dfrac {t^3}{3}\times e^{\log (t)}\times \dfrac {1}{t}dt$
$k=e^{\log (t)}\dfrac {t^3}{3} \displaystyle \int \dfrac {t^2}{3} e^{\log (t)}dt$
$k=e^{\log (t)}\dfrac {t^3}{3}-\dfrac {1}{3}k$
$\dfrac {4k}{3}=e^{\log (t)}\dfrac {t^3}{3}$
$k=\dfrac {t^3}{4}e^{\log (t)}$

$L=\displaystyle \int e^{\log (t)}dt$
$L=e^{\log (t)}xt-\displaystyle \int t\times e^{\log (t)}\times \dfrac {1}{t}dt$
$L=t\ e^{\log t}-L$
$\Rightarrow \ 2L=te^{\log t}$
$\Rightarrow \ L=\dfrac {1}{2}[t\ e^{\log t}]$

$\therefore \ I=A-B=\dfrac {1}{2}t e^{\log t}-\dfrac {1}{4}t^3 e^{\log t}$
$=\dfrac {t^{\log t}}{2} \left [1-\dfrac {t^2}{2}\right]$
$\dfrac {\sin x}{2}e^{\log (\sin x)} \left [1-\dfrac {\sin^2x}{2}\right]$
$\dfrac {\sin x}{2}e^{\log (\sin x)} \left [\dfrac {1+1-\sin^2x}{2}\right]$
$I=\dfrac {\sin x}{4}e^{\log (\sin x)} [1+\cos^2x]$
$\therefore \ \displaystyle \int \cos^3 x e^{\log (\sin x)}dx=\dfrac {\sin x}{4}[1+\cos^2x]e^{\log (\sin x)}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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