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Question Number 115030 by bobhans last updated on 23/Sep/20

$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}\sqrt{\sec x-\cos x}dx=?$$

Answered by bemath last updated on 23/Sep/20

$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}\sqrt{\frac{1-\cos {}^{2}x}{\cos x}}dx=\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\mid \sin x\mid }{\sqrt{\cos x}}dx$$$$=-\underset{-\frac{\pi }{2}}{\stackrel{0}{\int }}\frac{\sin x}{\sqrt{\cos x}}dx+\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sin x}{\sqrt{\cos x}}dx$$$$=\underset{-\frac{\pi }{2}}{\stackrel{0}{\int }}\frac{d\left(\cos x\right)}{\sqrt{\cos x}}-\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{d\left(\cos x\right)}{\sqrt{\cos x}}$$$$=2{\left[\sqrt{\cos x}\right]}_{-\frac{\pi }{2}}^0-2\left[\sqrt{\cos x}\right]{}_0^{\frac{\pi }{2}}$$$$=2-2(0-1)=4$$

Answered by mathmax by abdo last updated on 23/Sep/20

$$\mathrm{A}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{\frac{1}{\mathrm{cosx}}-\mathrm{cosx}}\mathrm{dx}\Rightarrow \mathrm{A}=2{\int }_0^{\frac{\pi }{2}}\sqrt{\frac{1-{\cos }^2\mathrm{x}}{\mathrm{cosx}}}\mathrm{dx}$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{\mid \mathrm{sinx}\mid }{\sqrt{\mathrm{cosx}}}\mathrm{dx}=2{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sinx}}{\sqrt{\mathrm{cosx}}}\mathrm{dx}=2{[-2\sqrt{\mathrm{cosx}}]}_0^{\frac{\pi }{2}}=2\left\{2\right\}=4$$

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