Question

integral of √1+sinx- √1-sinx

Answer 1

Step-by-step explanation:

$\green{ \large{ \sf \: SOLUTION}}$

$\sf \int \{\sqrt{1 + \sin(x) } - \sqrt{1 - \sin(x) } \} dx\\ \\ \sf \int \sqrt{1 + \cos \bigg( \frac{\pi}{2} - x \bigg) } dx \: \: - \int\sqrt{1 - \cos \bigg( \frac{\pi}{2} - x \bigg) } dx$

$\sf \large \: FORMULA \\ \star \sf \: 1 + \cos( \theta) = 2 \cos {}^{2} ( \frac{ \theta}{2} ) \\ \\ \star \sf \: 1 - \cos( \theta) = 2 \sin {}^{2} ( \frac{ \theta}{2} )$

$\sf \int \sqrt{1 + \cos \bigg( \frac{\pi}{2} - x \bigg) } dx \: \: - \int\sqrt{1 - \cos \bigg( \frac{\pi}{2} - x \bigg) } dx \\ \\ \sf = \int \sqrt{2 \cos {}^{2} \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) } dx \: - \int \sqrt{2 \sin {}^{2} \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) } dx \\ \\ \sf = \int \sqrt{2} \cos \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) dx - \int \sqrt{2} \sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \\ \\ \sf = \sqrt{2} \bigg \{ \int \cos \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) dx - \int \sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \bigg \}$

$\sf \: let \: \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) = t\\ \\ \sf - \frac{dx}{2} = dt \\ \\ \sf \: dx \: = - 2dt$

$\sf \sqrt{2} \bigg \{ \int \cos \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) dx - \int \sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \bigg \} \\ \\ = \sf \: \sqrt{2} \bigg \{ \int \cos(t) ( - 2dt) - \int \ \sin(t) ( - 2dt) \bigg \} \\ \\ = \sf \: 2 \sqrt{2} \bigg( - \int \cos(t) dt + \int \sin(t) dt \bigg) \\ \\ \sf \: = 2 \sqrt{2} \bigg( - \sin(t) + ( - \cos(t) ) \bigg) \\ \\ = \sf \: - 2 \sqrt{2} \bigg( \sin \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) + \cos \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg) \bigg)$