On an integral April 13, 2019 Tolaso Evaluate the integral Solution Successively we have: We used the simple observation that Read more They might interest you:Exponential fractional partTrigonometric inequalityRoot inequalityMaximum of a functionA logarithmic sin integralAn infinite productSeries with cosine integralOn a nested sin sequenceA Riemann sum IV
![\[\mathcal{J} = \int_{0}^{\pi/4} \left ( \frac{x}{x \sin x + \cos x} \right )^2 \, \mathrm{d}x\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-94db9d805f0614ecddbd6b09267c4999_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \mathcal{J} &=\int_{0}^{\pi/4} \left ( \frac{x}{x \sin x + \cos x} \right )^2 \, \mathrm{d}x \\ &=-\int_{0}^{\pi/4} \frac{-x \cos x}{\left ( x \sin x + \cos x \right )^2} \cdot \frac{x}{\cos x} \, \mathrm{d}x \\ &=-\int_{0}^{\pi/4} \left ( \frac{1}{x \sin x+ \cos x} \right )' \frac{x}{\cos x} \, \mathrm{d}x \\ &=-\left [ \frac{1}{x\sin x + \cos x } \cdot \frac{x}{\cos x} \right ]_{0}^{\pi/4} + \int_{0}^{\pi/4} \frac{1}{x \sin x + \cos x} \cdot \left ( \frac{x}{\cos x} \right )' \, \mathrm{d}x \\ &= -\frac{2\pi}{\pi+4} + \int_{0}^{\pi/4} \frac{1}{x \sin x + \cos x} \cdot \frac{x \sin x+ \cos x}{\cos^2 x} \, \mathrm{d}x \\ &= -\frac{2\pi}{\pi+4} + \int_{0}^{\pi/4} \frac{\mathrm{d}x}{\cos^2 x}\\ &= -\frac{2\pi}{\pi+4} + \left [ \tan x \right ]_{0}^{\pi/4} \\ &= 1-\frac{2\pi}{\pi+4} \\ &=\frac{4-\pi}{4+\pi} \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-158168ec00108c0f3b76716b6631106a_l3.png "Rendered by QuickLaTeX.com")
![\[\left ( \frac{1}{x \sin x + \cos x} \right ) ' = - \frac{x \cos x}{\left ( x \sin x + \cos x \right )^2}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-22c75c93405e87ec6cb730d4372fb0e4_l3.png "Rendered by QuickLaTeX.com")