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Question Number 78334 by john santu last updated on 16/Jan/20

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}[\cos {}^2\left(\cos x\right)+{\sin }^2\left(\sin x\right)]dx$$

Commented by MJS last updated on 16/Jan/20

$$\mathrm{we}\mathrm{cannot}\mathrm{solve}\mathrm{this}\mathrm{integral}\mathrm{but}\mathrm{we}\mathrm{can}$$$$\mathrm{approximate}f\left(x\right)={\cos }^2\left(\cos x\right)+{\sin }^2\left(\sin x\right)$$$$\mathrm{by}\mathrm{assuming}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{shifted}\mathrm{cosine}$$$$f\left(x\right)\sim g\left(x\right)=1-{\sin }^{2}1\cos 2x=$$$$\mathrm{plotting}\mathrm{both}\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{difference}$$$$\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{for}0⩽x⩽\frac{\pi }{4}\mathrm{is}\mathrm{the}\mathrm{same}$$$$\mathrm{for}\frac{\pi }{4}⩽x⩽\frac{\pi }{2}\mathrm{with}\mathrm{reversed}\mathrm{sign}\Rightarrow$$$$\Rightarrow \underset{0}{\stackrel{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}g\left(x\right)dx=\frac{\pi }{2}$$

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