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Question Number 32675 by Cheyboy last updated on 31/Mar/18

Commented by caravan msup abdo. last updated on 31/Mar/18

$$wehave1+cos\left(4x\right)=2{cos}^2\left(2x\right)\Rightarrow$$$$I={\int }_0^{\frac{\pi }{2}}\sqrt{1+cos\left(4x\right)}dx$$$$=\sqrt{2}{\int }_0^{\frac{\pi }{2}}\mid cos\left(2x\right)\mid dx$$$$=\sqrt{2}{\int }_0^{\frac{\pi }{4}}cos\left(2x\right)dx-\sqrt{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}cos\left(2x\right)dx$$$$=\frac{\sqrt{2}}{2}{\left[sin\left(2x\right)\right]}_0^{\frac{\pi }{4}}-\frac{\sqrt{2}}{2}{\left[sin\left(2x\right)\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$$$$=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}(-1)=\sqrt{2}.$$

Answered by MJS last updated on 31/Mar/18

$$\cos 2x={2\cos }^{2}x-1$$$${2\cos }^{2}2x-1={8\cos }^{4}x-{8\cos }^{2}x+1\Rightarrow$$$$\Rightarrow 1+\cos 4x=2({4\cos }^{4}x-{4\cos }^{2}x+1)=$$$$=2{({2\cos }^{2}x-1)}^2={2\cos }^{2}2x$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\sqrt{2}\mid \cos 2x\mid dx=2\sqrt{2}\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\cos 2xdx=$$$$=2\sqrt{2}\frac{\sin 2x}{2}\underset{0}{\stackrel{\frac{\pi }{4}}{\mid }}=\sqrt{2}\sin 2x\underset{0}{\stackrel{\frac{\pi }{4}}{\mid }}=\sqrt{2}$$

Commented by Cheyboy last updated on 31/Mar/18

$$Thankyousirbttheupper$$$$limitir\pi /2$$

Commented by MJS last updated on 31/Mar/18

$$\mathrm{I}\mathrm{changed}\mathrm{the}\mathrm{upper}\mathrm{limit}$$$$\mathrm{because}\mathrm{I}\mathrm{used}\cos 2x\mathrm{instead}$$$$\mathrm{of}\mid \cos 2x\mid .\cos 2x\mathrm{is}⩾0\mathrm{in}[0;\frac{\pi }{4}]$$$$\mathrm{but}⩽0\mathrm{in}[\frac{\pi }{4};\frac{\pi }{2}],\mathrm{so}\sqrt{2}\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\cos 2xdx=0$$$$\mathrm{but}\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\cos 2xdx=-\underset{\frac{\pi }{4}}{\stackrel{\frac{\pi }{2}}{\int }}\cos 2xdx\mathrm{so}$$$$\sqrt{2}\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\mid \cos 2x\mid dx=2\sqrt{2}\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\cos 2xdx$$$$$$

Commented by Cheyboy last updated on 31/Mar/18

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{r}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{z}\mathit{w}\mathit{e}\mathit{l}\mathit{l}\mathit{u}\mathit{n}\mathit{d}\mathit{e}\mathit{r}\mathit{s}\mathit{t}\mathit{o}\mathit{o}\mathit{d}$$

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