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Integrate the function\[\int\cos^3x\;e^{log\sin x}dx\]

$\begin{array}{1 1} -\large \frac{\cos ^4x}{4}+c \\ -\large \frac{\sin ^4x}{4}+c \\ -\large \frac{\log \mid cos ^4x\mid}{4}+c \\-\large \frac{\log \mid sin^4x\mid}{4}+c\end{array} $

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  • (i)$ e^{logx}=x$
  • (ii)$\int \frac{x^n}{1}=\frac{x^{n+1}}{n+1}$
  • (iii) If an integral function $f(x)=t,$ then $f'(x)dx=dt$. Hence $\int f(x)dx=\int tdt$
Given $I=\int\cos^3x\;e^{log\sin x}dx$
Put $\cos x=t, $ on differentiating w.r.t. x we get $-\sin x dx=dt\qquad=>\sin x dx=-dt$
Substituting for t and dt
$I=\int t^3 (-dt)$
$=-\int t^3dt$
On integrating we get
Substituting for t we get
$I=-\bigg(\frac{\cos ^4x}{4}\bigg)+c$



answered Feb 17, 2013 by meena.p
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