Question

integrate :sin^2x / (sinx + cosx) limit - 0 to Pie/2

Answer 1

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Answer 2

Given:

$\bold{\int_{0}^{\frac{\pi}{2}} \frac{sin^2x}{(sinx + cosx)}}$

To find:

integrate the value.

Solution:

Formula:

$\int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$\int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx$

$\Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{sin^2x}{(sinx + cosx)}$

$\Rightarrow \int _0^{\frac{\pi }{2}}\frac{\frac{1-\cos (2x)}{2}}{\cos(x)+\sin(x)}dx$

$\Rightarrow \frac{1}{2}\cdot \int _0^{\frac{\pi }{2}}\frac{1-\cos (2x)}{\cos (x)+\sin (x)}dx$

$\Rightarrow \frac{1}{2}(\int _0^{\frac{\pi }{2}}\frac{1}{\cos(x)+\sin(x)}dx-\int _0^{\frac{\pi }{2}}\frac{\cos (2x)}{\cos (x)+\sin (x)}dx)$

$\\ \int _0^{\frac{\pi }{2}}\frac{1}{\cos(x)+\sin(x)}dx = (-\frac{1}{\sqrt{2}}(\ln (-\frac{1}{\sqrt{2}}+1)-\ln(\frac{1}{\sqrt{2}}+1)))\\\\ \int _0^{\frac{\pi }{2}}\frac{\cos (2x)}{\cos (x)+\sin (x)}dx) = 0$

$\Rightarrow \frac{1}{2}(-\frac{1}{\sqrt{2}}(\ln (-\frac{1}{\sqrt{2}}+1)-\ln(\frac{1}{\sqrt{2}}+1))-0))\\\\\Rightarrow -\frac{1}{2\sqrt{2}}\left(\ln \left(-\frac{1}{\sqrt{2}}+1\right)-\ln \left(\frac{1}{\sqrt{2}}+1\right)\right)$