JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 8)
$$
\int\limits_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x \text { is equal to }
$$
$$10(\pi+4)$$
$$10(\pi+2)$$
$$20(\pi-2)$$
$$20(\pi+2)$$
Explanation
$$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $$
$$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $$
$$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $$
$$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$$
$$ = 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)$$
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