JEE MAIN - Mathematics (2004 - No. 30)
The value of $$I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx} $$ is
$$3$$
$$1$$
$$2$$
$$0$$
Explanation
$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$$
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
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
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