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Question Number 187890 by Rupesh123 last updated on 23/Feb/23

Answered by aleks041103 last updated on 23/Feb/23

$$I=\underset{-1}{\stackrel{1}{\int }}\frac{4\sqrt{1-x^2}}{1+2^x}dx$$$$u=-x,du=-dx$$$$I={\int }_1^{-1}\frac{4\sqrt{1-{(-u)}^2}}{1+2^{-u}}(-du)=$$$$={\int }_{-1}^1\frac{2^u}{1+2^u}4\sqrt{1-u^2}du=$$$$={\int }_{-1}^{1}4\sqrt{1-u^2}du-{\int }_{-1}^1\frac{4\sqrt{1-u^2}}{1+2^u}du$$$$\Rightarrow I=2\underset{-1}{\stackrel{1}{\int }}\sqrt{1-x^2}dx=\pi$$$$\Rightarrow I=\pi$$

Commented by Rupesh123 last updated on 23/Feb/23

Perfect ��

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