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
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