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The value of 1=\int_{0}^{\frac{\pi}{2}}\frac{(\sin x +\cos x)^{2}}{\sqrt{(1+sin2x)}}dxis

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lower and upper limit -

\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}

                = F\left ( b \right )-F\left ( a \right )

 

- wherein

Where a is lower and b is upper limit.

 

 I=\int_{0}^{\frac{\pi }{2}}\frac{(sinx+cosx)^2}{\sqrt{1+sin2x}} dx

I=\int_{0}^{\frac{\pi }{2}}\frac{(sinx+cosx)^2}{{sinx+cosx}} dx

I=\int_{0}^{\frac{\pi }{2}}{{(sinx+cosx)}} dx                                          [1+sin2x = (sinx +cosx)^2]

= -cosx +sin x|^{\frac{\Pi }{2}}_0 =[-cos\frac{\Pi }{2}+sin\frac{\Pi }{2}+cos 0-sin0] =2


Option 1)

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