Mathematics

$$\int_\limits{0}^{\pi/2}\frac{sinx - cosx}{1+ sinx cosx}dx$$


SOLUTION

$$\int_{0}^{(\frac{\pi}{2})}(\frac{sin(\frac{\pi}{2}-x)-cosx(\frac{\pi}{2}-x)}{1+sin(\frac{\pi}{2}-x) cosx(\frac{\pi}{2}-x)})dx\\=\int_{0}^{(\frac{\pi}{2})}(\frac{cosx-sinx}{1+cosx sinx})dx \\=-\int_{0}^{(\frac{\pi}{2})}(\frac{cosx-sinx}{1+cosx sinx})dx\\ \therefore I =-I\\ \therefore 2I=0 \>then \>I =0$$

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Subjective Medium Published on 17th 09, 2020
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