Question Number 63509 by mathmax by abdo last updated on 05/Jul/19
$${calculate}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$
Commented by Prithwish sen last updated on 05/Jul/19
![2∫_0 ^1 ((√(1+x^2 )) − (√(1−x^2 ))) dx = 2[(x/2)(√(1+x^2 )) + (1/2)ln∣x+(√(1+x^2 ))∣ −(x/2)(√(1−x^2 )) − (1/2) sin^(−1) x ]_0 ^1 = 2[(1/2)(√2) +(1/2)ln∣1+(√2)∣− (π/4) ] = ((√2) − (π/(2 )) ) + ln∣1+(√2)∣ please check.](https://www.tinkutara.com/question/Q63520.png)
$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} \:}\:−\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)\:\mathrm{dx} \\ $$$$=\:\mathrm{2}\left[\frac{\mathrm{x}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mid\:−\frac{\mathrm{x}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \mathrm{x}\:\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid−\:\frac{\pi}{\mathrm{4}}\:\right]\:=\:\left(\sqrt{\mathrm{2}}\:−\:\frac{\pi}{\mathrm{2}\:}\:\right)\:+\:\mathrm{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$
Answered by MJS last updated on 05/Jul/19
![−∫_(−1) ^1 (√(1−x^2 ))dx=−(π/2) ∫_(−1) ^1 (√(1+x^2 ))dx=(1/2)[x(√(1+x^2 ))+ln (x+(√(1+x^2 )))]_(−1) ^1 =(√2)+ln (1+(√2)) ∫_(−1) ^1 ((√(1+x^2 ))−(√(1−x^2 )))dx=(√2)+ln (1+(√2)) −(π/2)](https://www.tinkutara.com/question/Q63526.png)
$$−\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=−\frac{\pi}{\mathrm{2}} \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} =\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx}=\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\pi}{\mathrm{2}} \\ $$