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

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

Toolbox:
• (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

$I=-\bigg[\frac{t^4}{4}\bigg]+c$

Substituting for t we get

$I=-\bigg(\frac{\cos ^4x}{4}\bigg)+c$