The value of I = _0^/2 ( x+ x)^2√1+2x,dx is:

Question

The value of $I = \displaystyle\int_0^{\pi/2} \dfrac{(\sin x+\cos x)^2}{\sqrt{1+\sin2x}}\,dx$ is:

Accepted Answer

$$I = \int\limits_0^{{\pi  \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$$

We know $$\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$$ So

$$I = \int\limits_0^{{\pi  \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}} dx$$

$$ = \int\limits_0^{{\pi  \over 2}} {\left( {\sin x + \cos x} \right)dx} $$

$$\left[ {} \right.$$  $$\sin x + \cos x > 0$$ $$\,\,if\,\,0 < x < {\pi  \over 2}$$ $$\left. {} \right]$$

or $$I = \left[ { - \cos x + \sin x} \right]_0^{{\pi  \over 2}} $$

= - cos $${\pi  \over 2}$$ + sin $${\pi  \over 2}$$ + cos 0 - sin 0

= - 0 + 1 + 1 - 0

= 2

Solutions · 1