What is f(x) = int xcotx^2 dx if f((5pi)/4) = 0 ?

May 5, 2018

I have tried , but it is complicated for ${x}^{2} = {\left(\frac{5 \pi}{4}\right)}^{2.}$

$f \left(x\right) = \frac{1}{2} \ln | \sin {x}^{2} | + c , w h e r e , c \approx 0.6623$

Explanation:

Here,

$f \left(x\right) = I = \int x \cot {x}^{2} \mathrm{dx} = \int \cot {x}^{2} \cdot x \mathrm{dx}$

Let, ${x}^{2} = u \implies 2 x \mathrm{dx} = \mathrm{du} \implies x \mathrm{dx} = \frac{1}{2} \mathrm{du}$

So,

$I = \int \cot u \cdot \frac{1}{2} \mathrm{du}$

$= \frac{1}{2} \ln | \sin u | + c , w h e r e , u = {x}^{2}$

$\implies f \left(x\right) = \frac{1}{2} \ln | \sin {x}^{2} | + c \ldots \to \left(A\right)$

Given that,

$f \left(\frac{5 \pi}{4}\right) = 0$

$\implies \frac{1}{2} \ln | \sin {\left(\frac{5 \pi}{4}\right)}^{2} | + c = 0$

$\implies 2 c = - \ln | \sin \left(\frac{25 {\pi}^{2}}{16}\right) |$

$\implies 2 c = \ln | \left(\frac{1}{\sin \left(\frac{25 {\pi}^{2}}{16}\right)}\right) |$

$\implies c = \frac{1}{2} \ln | \left(\frac{1}{\sin \left(\frac{25 {\pi}^{2}}{16}\right)}\right) | \ldots \to \left(1\right)$

Now,

(25xxpi^2)/16 ~~15.42....

$\sin \left(\frac{25 {\pi}^{2}}{16}\right) \approx 0.2659 \ldots \in \left[- 1 , 1\right]$

$\frac{1}{\sin \left(\frac{25 {\pi}^{2}}{16}\right)} \approx 3.7606 \ldots$

$\ln | \frac{1}{\sin \left(\frac{25 {\pi}^{2}}{16}\right)} | \approx 1.3245 \ldots$

$\frac{1}{2} \cdot \ln | \frac{1}{\sin \left(\frac{25 {\pi}^{2}}{16}\right)} | \approx 0.6623$

Hence, from $\left(1\right)$

$c \approx$0.6623

Thus,

$f \left(x\right) = \frac{1}{2} \ln | \sin {x}^{2} | + c , w h e r e , c \approx 0.6623$