Mathematics


$$\displaystyle \int_{0}^{\pi /2}\frac{\sin^{2}x}{\sin x+\cos x}dx=$$


ANSWER

$$\displaystyle \frac{1}{\sqrt{2}}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$$


SOLUTION

$$I=\int_{0}^{\dfrac{\pi}{2}}\dfrac{sin^2  x}{sin  x +  cos  x}dx$$
$$I=\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos^2}{sin  x +  cos  x}dx$$
$$2I= \int_{0}^{\dfrac{\pi}{2}}\dfrac{1}{sin  x +  cos  x}dx$$
$$2I=\int_{0}^{\dfrac{\pi}{2}}\dfrac{\left (  1+ tan^2 \dfrac{x}{2}\right )}{-tan^2 \dfrac{x}{2}+ 2 tan \dfrac{x}{2}+1} dx$$
$$2I =\int_{0}^{\dfrac{\pi}{2}}\dfrac{2 d  tan  \dfrac{x}{2}}{(\sqrt{2})^2 \left ( tan \dfrac{x}{2} -1\right )^2}$$
$$=\dfrac{1}{2\sqrt{2}} log \left | \dfrac{\sqrt{2}+ tan^n (\dfrac{x}{2})-1}{\sqrt{2}(tan \dfrac{x}{2})+1}  \right |$$
$$=\dfrac{1}{2\sqrt{2}} log \left | \dfrac{ tan^n (\dfrac{x}{2})+\sqrt{2}-1}{(\sqrt{2}+1)tan \dfrac{x}{2}}  \right |\int_{0}^{\dfrac{\pi}{4}}$$
$$=\dfrac{1}{2}\sqrt{2} log ((\sqrt{2}+1)^2) = \dfrac{1}{\sqrt{2}} log (\sqrt{2}+1)$$

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Single Correct Hard Published on 17th 09, 2020
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