$Find the average value of the function $\frac{\sin x}{1+\cos^2 x}$ on the interval $[0,1]$$

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$Find the average value of the function $\frac{\sin x}{1+\cos^2 x}$ on the interval $[0,1]$$

Find the average value of the function
$$f(x)=\frac{\sin x}{1+\cos^2 x}$$
on the interval $[0,1]$.

Calculus Integrals Trigonometric Functions

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Let $u=\cos x$, then $du=-\sin x dx$
\[\int_0^1\frac{\sin x}{1+\cos ^2 x}dx=\int_{1}^{\cos 1}\frac{-du}{1+u^2}du=\int_{\cos 1}^{ 1}\frac{du}{1+u^2}du\]
\[=\arctan (u)|_{\cos 1}^{1}=\arctan (1)-\arctan (\cos (1)).\]

So the average value is
\[=\frac{\int_0^1\frac{\sin x}{1+\cos ^2 x}dx}{1-0}=\arctan (u)|_{\cos 1}^{1}=\arctan (1)-\arctan (\cos (1)).\]

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$Find the average value of the function $\frac{\sin x}{1+\cos^2 x}$ on the interval $[0,1]$$

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