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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=∫_(−1) ^1 ((4(√(1−x^2 )))/(1+2^x ))dx  u=−x, du=−dx  I=∫_1 ^(−1) ((4(√(1−(−u)^2 )))/(1+2^(−u) ))(−du)=  =∫_(−1) ^1 (2^u /(1+2^u ))4(√(1−u^2 ))du=  =∫_(−1) ^1 4(√(1−u^2 ))du−∫_(−1) ^1 ((4(√(1−u^2 )))/(1+2^u ))du  ⇒I=2∫_(−1) ^1 (√(1−x^2 ))dx=π  ⇒I=π

$${I}=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{4}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{2}^{{x}} }{dx} \\ $$$${u}=−{x},\:{du}=−{dx} \\ $$$${I}=\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\mathrm{4}\sqrt{\mathrm{1}−\left(−{u}\right)^{\mathrm{2}} }}{\mathrm{1}+\mathrm{2}^{−{u}} }\left(−{du}\right)= \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{2}^{{u}} }{\mathrm{1}+\mathrm{2}^{{u}} }\mathrm{4}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }{du}= \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{4}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{4}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{2}^{{u}} }{du} \\ $$$$\Rightarrow{I}=\mathrm{2}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\pi \\ $$$$\Rightarrow{I}=\pi \\ $$

Commented by Rupesh123 last updated on 23/Feb/23

Perfect ��

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