Question

$integration \sqrt{1 + \sin(2x) dx }$
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Answer 1

Solution :

$\mathsf{\int \sqrt{(1+sin2x)}\:dx}$

$\small{\mathsf{=\int \sqrt{(sin^{2}x+cos^{2}x+2\:sinx\:cosx)}\:dx}}$

$\mathsf{=\int \sqrt{(sinx+cosx)^{2}}\:dx}$

$\mathsf{=\int (sinx+cosx)\:dx}$

$\mathsf{=\int sinx\:dx+\int cosx\:dx}$

$\mathsf{=-cosx+sinx+C}$

where C is integral constant

 $\boxed{\small{\mathsf{\int \sqrt{1+sin2x}\:dx=-cosx+sinx+C}}}$

which is the required integral

Answer 2

$\bf \huge \underline{ \: \: \: \: Solution \: \: \: \: \: : } \\ \\ \\ \\ \\ \mathbf{ \int \sqrt{1 + \sin{2x}} \: dx} \\ \\ \to \mathbf{ \int \sqrt{({ \sin}^{2}x +{ \cos}^{2}x + 2 \sin x \cos x} ) \: dx} \\ \\ \to \int \bf ( \sin x + \cos x) \: dx \\ \\ \to \int \bf \sin x \: dx + \int \cos x \: dx \\ \\ \to \boxed{\boxed{\bf - \cos x + \sin x + C}} \\ \\ \\ \\ \sf where \: C \: is \: an \: integral \: constant.$