Complicated definite integral

$I = \int_0^1 \! \dfrac{cos(x) ln(cos(x)) + xsin(x)}{x^2 cos(x)} \ \mathrm{d}x = \int_0^1 \! \dfrac{ln(cos(x))}{x^2} \ \mathrm{d}x + \int_0^1 \! \dfrac{sin(x)}{x cos(x)} \ \mathrm{d}x$

Using partial integration on the first integral , with $u = ln(cos(x)) \hspace{2mm} \text{and} \hspace{2mm} v'= \dfrac{1}{x^2} \Rightarrow v = \dfrac{-1}{x}$ ,

we find for $I$ :

$I = \left . \dfrac{-ln(cos(x))}{x} \right|_0^1 - \int_0^1 \! \dfrac{sin(x)}{x cos(x)} + \int_0^1 \! \dfrac{sin(x)}{x cos(x)} = \left . \dfrac{-ln(cos(x))}{x} \right|_0^1 .$

Now plugging in 1, gives us -ln(cos(1)), while plugging in 0, gives us $\frac{0}{0}$

Using L'Hôpital's rule on this undetermined expression, gives us $tan(x)$ , which is equal to zero when x equals zero.

So $I = -ln(cos(1))$ , $f(x) = cos(x)$ and $cos(\pi) = -1$ .

If we ignore the limits in the definite integral above, the indefinite integral cannot be easily found using elementary functions. We can use differentiation under the integral sign to evaluate the definite integral without having to evaluate the indefinite integral.

$\mathrm{I}(t) = \int_0^1 \frac{\cos\left(tx\right)\ln\left(\cos\left(tx\right)\right)+tx\sin\left(tx\right)}{x^{2}\cos\left(tx\right)}\,\mathrm{d}x = \int_0^1 \frac{\ln\left(\cos\left(tx\right)\right)}{x^{2}}+\frac{t\tan\left(tx\right)}{x}\,\mathrm{d}x$

$\mathrm{I}'(t) = \frac{\mathrm d}{\mathrm dt} \int_0^1 \frac{\ln\left(\cos\left(tx\right)\right)}{x^{2}}+\frac{t\tan\left(tx\right)}{x}\,\mathrm{d}x = \int_0^1 \frac{\mathrm \delta}{\mathrm \delta t} \left( \frac{\ln\left(\cos\left(tx\right)\right)}{x^{2}}+\frac{t\tan\left(tx\right)}{x}\right)\,\mathrm{d}x$

$\mathrm{I}'(t) = \int_0^1 -\frac{\tan(tx)}{x} + \frac{\tan(tx)}{x} +t\sec^{2}(tx)\,\mathrm{d}x = \int_0^1 t\sec^{2}(tx)\,\mathrm{d}x$

Now we can evaluate this integral with respect to $x$ and find an expression for $\mathrm I'(t)$ entirely in terms of $t$ .

$\mathrm{I}'(t) = \int_0^1 t\sec^{2}(tx)\,\mathrm{d}x = \tan{tx}\,\Big|_{x=0}^{x=1} = \tan{t}$

Now that we have found $\mathrm I'(t)$ we can determine an expression for $\mathrm I(t)$ .

$\mathrm{I}(t) = \int \tan{t}\,\mathrm{d}t = -\ln(\cos(t)) + c = \ln(\sec(t)) + c$

To solve for the constant of integration, we go back to the original integral. $\mathrm{I}(0) = \int_0^1 \frac{\ln\left(\cos\left(0\right)\right)}{x^{2}}+\frac{0\tan\left(0\right)}{x}\,\mathrm{d}x = 0$ . Substituting these into the equation for $\mathrm{I}(t)$ shows that $c=0$ and hence $\mathrm{I}(t) = \ln(\sec(t))$ . To find the value of the original definite integral, substitute $t=1$ into $\mathrm{I}(t)$ to obtain $\boxed{\ln\left(\sec(1)\right)}$ . Hence $f(x) = \sec(x)$ and $\boxed{f(\pi) = -1}$ .