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1-3-tan-1-x-x-2-1-tan-1-x-2-1-x-dx-

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Question Number 83075 by mhmd last updated on 27/Feb/20
∫_(−1 ) ^3 (tan^(−1) (x/(x^2 +1)) + tan^(−1) ((x^2 +1)/x))dx =
$$\underset{−\mathrm{1}\:\:} {\overset{\mathrm{3}} {\int}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\right){dx}\:= \\ $$
Commented by mathmax by abdo last updated on 27/Feb/20
we have arctan(u)+arctan((1/u))=(π/2) if u>0 and arctanu +arctan((1/u))=−(π/2) if u<0 ⇒ ∫_(−1) ^3 (arctan((x/(x^2 +1)))+arctan(((x^2 +1)/x)))dx =∫_(−1) ^0 (arctan((x/(x^2 +1)))+arctan(((x^2 +1)/x)))dx +∫_0 ^3 (arctan((x/(x^2 +1)))+arctan(((x^2 +1)/x)))dx =−(π/2)∫_(−1) ^0 dx +(π/2)∫_0 ^3 dx =−(π/2)(0+1)+(π/2)×3 =−(π/2)+((3π)/2) =π
$${we}\:{have}\:{arctan}\left({u}\right)+{arctan}\left(\frac{\mathrm{1}}{{u}}\right)=\frac{\pi}{\mathrm{2}}\:{if}\:{u}>\mathrm{0}\:{and} \\ $$$${arctanu}\:+{arctan}\left(\frac{\mathrm{1}}{{u}}\right)=−\frac{\pi}{\mathrm{2}}\:{if}\:{u}<\mathrm{0}\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{3}} \left({arctan}\left(\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+{arctan}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\right)\right){dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \left({arctan}\left(\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right)+{arctan}\left(\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{3}} \left({arctan}\left(\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right)+{arctan}\left(\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}}\right)\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{0}} \:{dx}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{3}} \:{dx}\:=−\frac{\pi}{\mathrm{2}}\left(\mathrm{0}+\mathrm{1}\right)+\frac{\pi}{\mathrm{2}}×\mathrm{3} \\ $$$$=−\frac{\pi}{\mathrm{2}}+\frac{\mathrm{3}\pi}{\mathrm{2}}\:=\pi \\ $$
Answered by mind is power last updated on 27/Feb/20
tan^− (x)+tan^− ((1/x))=(π/2)sign(x) =Σ_(k=0) ^3 ∫_(k−1) ^k {tan^− ((x/(1+x^2 )))+tan^− (((x^2 +1)/x))}dx =∫_(−1) ^0 −(π/2)dx+Σ_(k=0) ^2 ∫_k ^(k+1) (π/2)dx =−(π/2)+((3π)/2)=π
$${tan}^{−} \left({x}\right)+{tan}^{−} \left(\frac{\mathrm{1}}{{x}}\right)=\frac{\pi}{\mathrm{2}}{sign}\left({x}\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\int_{{k}−\mathrm{1}} ^{{k}} \left\{{tan}^{−} \left(\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+{tan}^{−} \left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\right)\right\}{dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} −\frac{\pi}{\mathrm{2}}{dx}+\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \frac{\pi}{\mathrm{2}}{dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}+\frac{\mathrm{3}\pi}{\mathrm{2}}=\pi \\ $$

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