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Question Number 38714 by maxmathsup by imad last updated on 28/Jun/18
calculate ∫_1 ^6 (((−1)^([x]) )/(1+x^2 [x]))dx
$${calculate}\:\:\:\int_{\mathrm{1}} ^{\mathrm{6}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\mathrm{1}+{x}^{\mathrm{2}} \left[{x}\right]}{dx} \\ $$
Commented by abdo mathsup 649 cc last updated on 29/Jun/18
I = Σ_(k=1) ^5 ∫_k ^(k+1) (((−1)^k )/(1+kx^2 ))dx =Σ_(k=1) ^5 (−1)^k ∫_k ^(k+1) (dx/(1+kx^2 )) changement x(√k)=t give ∫_k ^(k+1) (dx/(1+kx^2 )) = ∫_(k(√k)) ^((k+1)(√k)) (1/(1+t^2 )) (dt/( (√k))) ⇒ I =Σ_(k=1) ^5 (((−1)^k )/( (√k))) ∫_(k(√k)) ^((k+1)(√k)) (dt/(1+t^2 )) =Σ_(k=1) ^5 (((−1)^k )/( (√k))) { arctan(k+1)(√k) −arctan(k(√k))} =−arctan(2) +(π/4) +(1/( (√2))){ arctan(3(√2))−arctan(2(√(2))) −(1/( (√3))){ arctan(4(√3))−arctan(3(√3))} +(1/2){ar4tan(10) −arctan(8)} −(1/( (√5))){arctan6(√(5))) −arctan(5(√5)).
$${I}\:=\:\sum_{{k}=\mathrm{1}} ^{\mathrm{5}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{1}+{kx}^{\mathrm{2}} }{dx} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{5}} \:\left(−\mathrm{1}\right)^{{k}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{kx}^{\mathrm{2}} }\:{changement} \\ $$$${x}\sqrt{{k}}={t}\:{give}\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{kx}^{\mathrm{2}} }\:=\:\int_{{k}\sqrt{{k}}} ^{\left({k}+\mathrm{1}\right)\sqrt{{k}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{{dt}}{\:\sqrt{{k}}}\:\Rightarrow \\ $$$${I}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{5}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\:\sqrt{{k}}}\:\int_{{k}\sqrt{{k}}} ^{\left({k}+\mathrm{1}\right)\sqrt{{k}}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{5}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\:\sqrt{{k}}}\:\left\{\:{arctan}\left({k}+\mathrm{1}\right)\sqrt{{k}}\:−{arctan}\left({k}\sqrt{{k}}\right)\right\} \\ $$$$=−{arctan}\left(\mathrm{2}\right)\:+\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\:{arctan}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)−{arctan}\left(\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.\right. \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\mathrm{4}\sqrt{\mathrm{3}}\right)−{arctan}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\left\{{ar}\mathrm{4}{tan}\left(\mathrm{10}\right)\:−{arctan}\left(\mathrm{8}\right)\right\}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left\{{arctan}\mathrm{6}\sqrt{\left.\mathrm{5}\right)}\:−{arctan}\left(\mathrm{5}\sqrt{\mathrm{5}}\right).\right. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18
∫_1 ^2 (((−1)^1 )/(1+1×x^2 ))+∫_2 ^3 (((−1)^2 )/(1+2x^2 ))+∫_3 ^4 (((−1)^3 )/(1+3x^2 ))+∫_4 ^5 (((−1)^4 )/(1+4x^2 ))+ ∫_5 ^6 (((−1)^5 )/(1+5x^2 )) =(−1)∣tan^(−1) x∣_1 ^2 +(1/2)×(1/(1/( (√2))))∣tan^(−1) ((x/(1/( (√2)))))∣_2 ^3 + (−1)^ ×(1/3)×(1/(1/( (√3))))∣tan^(−1) ((x/(1/( (√3)))))∣_3 ^4 + (1/4)×(1/(1/( (√4))))∣tan^(−1) ((x/(1/2)))∣_4 ^5 +(−1)×(1/5)×(1/(1/( (√5))))∣(tan^(−1) (x/((1(/))/( (√5)))))∣_5 ^6 =(tan^(−1) 1−tan^(−1) 2)+(1/( (√2)))(tan^(−1) 3(√2) −tan^(−1) 2(√(2))) −(1/( (√3)))(tan^(−1) 4(√3) −tan^(−1) 3(√3) )+(1/2)(tn^(−1) 10−tn_ ^(−1) 8 −(1/( (√5)))(tan^(−1) 6(√( 5)) −tan^(−1) 5(√5) ) =tan^(−1) (((−1)/3))+(1/( (√2)))tan^(−1) ((((√2) )/(13)))−(1/( (√3) ))tan^(−1) (((3(√3) )/(37))) +(1/2)tan^(−1) ((2/(81)))−(1/( (√5) ))tan^(−1) ((((√5) )/(151)))
$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{\mathrm{1}+\mathrm{1}×{x}^{\mathrm{2}} }+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\int_{\mathrm{3}} ^{\mathrm{4}} \frac{\left(−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }+\int_{\mathrm{4}} ^{\mathrm{5}} \frac{\left(−\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }+ \\ $$$$\:\:\int_{\mathrm{5}} ^{\mathrm{6}} \frac{\left(−\mathrm{1}\right)^{\mathrm{5}} }{\mathrm{1}+\mathrm{5}{x}^{\mathrm{2}} } \\ $$$$=\left(−\mathrm{1}\right)\mid{tan}^{−\mathrm{1}} {x}\mid_{\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\right)\mid_{\mathrm{2}} ^{\mathrm{3}} + \\ $$$$\:\left(−\mathrm{1}\right)^{} ×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right)\mid_{\mathrm{3}} ^{\mathrm{4}} + \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)\mid_{\mathrm{4}} ^{\mathrm{5}} +\left(−\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}}\mid\left({tan}^{−\mathrm{1}} \frac{{x}}{\frac{\mathrm{1}\frac{}{}}{\:\sqrt{\mathrm{5}}}}\right)\mid_{\mathrm{5}} ^{\mathrm{6}} \\ $$$$=\left({tan}^{−\mathrm{1}} \mathrm{1}−{tan}^{−\mathrm{1}} \mathrm{2}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({tan}^{−\mathrm{1}} \mathrm{3}\sqrt{\mathrm{2}}\:−{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right. \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({tan}^{−\mathrm{1}} \mathrm{4}\sqrt{\mathrm{3}}\:−{tan}^{−\mathrm{1}} \mathrm{3}\sqrt{\mathrm{3}}\:\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({tn}^{−\mathrm{1}} \mathrm{10}−{tn}_{} ^{−\mathrm{1}} \mathrm{8}\right. \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left({tan}^{−\mathrm{1}} \mathrm{6}\sqrt{\:\mathrm{5}}\:\:−{tan}^{−\mathrm{1}} \mathrm{5}\sqrt{\mathrm{5}}\:\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\:}{\mathrm{13}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:}{\mathrm{37}}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{81}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}\:}{\mathrm{151}}\right) \\ $$

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