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Question Number 202212 by Calculusboy last updated on 22/Dec/23

Commented by BOYQOBILOV last updated on 23/Dec/23

Answered by shunmisaki007 last updated on 23/Dec/23

$$I=\underset{0}{\stackrel{\pi }{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx+\underset{\frac{\pi }{2}}{\stackrel{\pi }{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx$$$$\mathrm{Consider}\underset{\frac{\pi }{2}}{\stackrel{\pi }{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx.$$$$\Rightarrow \mathrm{Let}u=\pi -x,du=-dx$$$$\Rightarrow \underset{\frac{\pi }{2}}{\stackrel{\pi }{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx$$$$=-\underset{\frac{\pi }{2}}{\stackrel{0}{\int }}\frac{1}{1+{\left(\sin (\pi -u)\right)}^{\tan (\pi -u)}}du$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1}{1+{\left(\sin \left(u\right)\right)}^{-\tan \left(u\right)}}du$$$$\Rightarrow I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx+\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}dx$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}+\frac{1}{1+{\left(\sin \left(x\right)\right)}^{-\tan \left(x\right)}}dx$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}+1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}}{1+{\left(\sin \left(x\right)\right)}^{\tan \left(x\right)}+{\left(\sin \left(x\right)\right)}^{-\tan \left(x\right)}+1}dx$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}dx$$$$\therefore I=\frac{\pi }{2}★$$

Commented by Calculusboy last updated on 23/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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