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Question-168366




Question Number 168366 by cortano1 last updated on 09/Apr/22
Answered by qaz last updated on 09/Apr/22
∫_(−π/2) ^(π/2) ((cos x)/(1−x+(√(x^2 +1))))dx=∫_(−π/2) ^(π/2) ((cos x)/(1+x+(√(x^2 +1))))dx  =(1/2)∫_(−π/2) ^(π/2) ((2(1+(√(x^2 +1)))cos x)/((1+(√(x^2 +1)))^2 −x^2 ))dx=(1/2)∫_(−π/2) ^(π/2) (((1+(√(x^2 +1)))cos x)/(1+(√(x^2 +1))))dx  =∫_0 ^(π/2) cos xdx=1
$$\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}=\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{cos}\:\mathrm{x}}{\left(\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \frac{\left(\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}\:\mathrm{xdx}=\mathrm{1} \\ $$
Commented by peter frank last updated on 09/Apr/22
thanks
$$\mathrm{thanks} \\ $$

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