Question Number 54070 by rahul 19 last updated on 28/Jan/19
$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right) \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right).\left(\mathrm{co}{t}^{−\mathrm{1}} \frac{{x}}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mid{x}\mid} }}\right){dx} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{10}\right)\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\left.\mathrm{3}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({cotx}\right)}{\left(\left(\mathrm{sin}{x}\right)^{\mathrm{2009}} +\left(\mathrm{cos}{x}\right)^{\mathrm{2009}} \right)^{\mathrm{2}} }.\left(\mathrm{sin2}{x}\right)^{\mathrm{2008}} {dx} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{4}{x}+\mathrm{10}}{\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}\right)^{\mathrm{2}} }\:{dx}. \\ $$
Commented by maxmathsup by imad last updated on 28/Jan/19
![4) ∫_0 ^2 ((4x+10)/((x^2 +5x +6)^2 )) =2 ∫_0 ^2 ((2x+5)/((x^2 +5x+6)^2 )) =[−2 (1/(x^2 +5x +6))]_0 ^2 =−2( (1/(20)) −(1/6)) =(1/3) −(1/(10)) =(7/(30)).](https://www.tinkutara.com/question/Q54083.png)
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{4}{x}+\mathrm{10}}{\left({x}^{\mathrm{2}} +\mathrm{5}{x}\:+\mathrm{6}\right)^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}{x}+\mathrm{5}}{\left({x}^{\mathrm{2}} \:+\mathrm{5}{x}+\mathrm{6}\right)^{\mathrm{2}} }\:=\left[−\mathrm{2}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{5}{x}\:+\mathrm{6}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=−\mathrm{2}\left(\:\frac{\mathrm{1}}{\mathrm{20}}\:−\frac{\mathrm{1}}{\mathrm{6}}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{10}}\:=\frac{\mathrm{7}}{\mathrm{30}}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
Commented by rahul 19 last updated on 30/Jan/19
$${No}\:{prof}.\:{U}\:{are}\:{wrong}… \\ $$$${It}\:{ain}'{t}\:{an}\:{odd}\:{function}\:{since} \\ $$$${cot}^{−\mathrm{1}} \left(−{x}\right)=\pi−\mathrm{cot}^{−\mathrm{1}} \left({x}\right)\:{for}\:{all}\:{x}\epsilon\mathbb{R}. \\ $$$${Ans}:\:\frac{\pi^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}. \\ $$$$\:{Indeed}\:{a}\:{very}\:{beautiful}\:{problem}. \\ $$
Commented by Meritguide1234 last updated on 30/Jan/19
Commented by maxmathsup by imad last updated on 30/Jan/19
$${let}\:\:{arccotan}\left({t}\right)={y}\:\Leftrightarrow{t}\:={cotan}\left({y}\right)\:=\frac{\mathrm{1}}{{tany}}\:\Rightarrow{tany}\:=\frac{\mathrm{1}}{{t}}\:\Rightarrow{y}\:={arctan}\left(\frac{\mathrm{1}}{{t}}\right) \\ $$$$=\overset{−} {+}\frac{\pi}{\mathrm{2}}−{arctan}\left({t}\right)\:\Rightarrow{arcotan}\left({t}\right)\:=\overset{−} {+}\frac{\pi}{\mathrm{2}}−{arctan}\left({t}\right)\:\:\Rightarrow \\ $$$${arcotan}\left(−{t}\right)=\overset{−} {+}\frac{\pi}{\mathrm{2}}\:+{arctan}\left({t}\right)\:\:{you}\:{are}\:{right}\:{sir}\:\:{the}\:{function}\:{is}\:{not}\:{odd}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
![2)∫_0 ^(π/2) ((sin^2 nθ)/(sin^2 θ))dθ [n=10] I_n =(1/2)∫_0 ^(π/2) ((1−cos2nθ)/(sin^2 θ))dθ I_n −I_(n−2) =(1/2)∫_0 ^(π/2) ((cos(2n−2)θ−cos2nθ)/(sin^2 θ))dθ =(1/2)∫_0 ^(π/2) ((2sin(2n−1)θsinθ)/(sin^2 θ))dθ =∫_0 ^(π/2) ((sin(2n−1)θ)/(sinθ))dθ J_n =I_n −I_(n−1) J_n −J_(n−1) =∫_0 ^(π/2) ((sin(2n−1)θ−sin(2n−3)θ)/(sinθ))dθ =∫_0 ^(π/2) ((2cos(2n−2)θsinθ)/(sinθ))dθ =2∣((sin(2n−2)θ)/(2n−2))∣_0 ^(π/2) =0 so J_n −J_(n−1) =0 J_n =J_(n−1) I_n −I_(n−1) =I_(n−1) −I_(n−2) I_n =2I_(n−1) −I_(n−2) ∫_0 ^(π/2) ((sin^2 (nθ))/(sin^2 θ))=I_n we have to find I_(10) I_n =2I_(n−1) −I_(n−2) [I_0 =0] I_2 =2I_1 −I_0 =2I_1 −0=2×(π/2) =π [I_1 =(π/2)] I_3 =2I_2 −I_1 =((3π)/2) I_4 =2π ..... thus on calculation (attached photo) we get I_(10) =5π A general formula thus obtained ∫_0 ^(π/2) ((sin^2 (nθ))/(sin^2 θ))dθ=n((π/2))=I_n](https://www.tinkutara.com/question/Q54112.png)
$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} {n}\theta}{{sin}^{\mathrm{2}} \theta}{d}\theta\:\:\:\:\left[{n}=\mathrm{10}\right] \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{cos}\mathrm{2}{n}\theta}{{sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$${I}_{{n}} −{I}_{{n}−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left(\mathrm{2}{n}−\mathrm{2}\right)\theta−{cos}\mathrm{2}{n}\theta}{{sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{sin}\left(\mathrm{2}{n}−\mathrm{1}\right)\theta{sin}\theta}{{sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left(\mathrm{2}{n}−\mathrm{1}\right)\theta}{{sin}\theta}{d}\theta \\ $$$${J}_{{n}} ={I}_{{n}} −{I}_{{n}−\mathrm{1}} \\ $$$${J}_{{n}} −{J}_{{n}−\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left(\mathrm{2}{n}−\mathrm{1}\right)\theta−{sin}\left(\mathrm{2}{n}−\mathrm{3}\right)\theta}{{sin}\theta}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{cos}\left(\mathrm{2}{n}−\mathrm{2}\right)\theta{sin}\theta}{{sin}\theta}{d}\theta \\ $$$$=\mathrm{2}\mid\frac{{sin}\left(\mathrm{2}{n}−\mathrm{2}\right)\theta}{\mathrm{2}{n}−\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{0} \\ $$$${so}\:{J}_{{n}} −{J}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${J}_{{n}} ={J}_{{n}−\mathrm{1}} \\ $$$${I}_{{n}} −{I}_{{n}−\mathrm{1}} ={I}_{{n}−\mathrm{1}} −{I}_{{n}−\mathrm{2}} \\ $$$${I}_{{n}} =\mathrm{2}{I}_{{n}−\mathrm{1}} −{I}_{{n}−\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \left({n}\theta\right)}{{sin}^{\mathrm{2}} \theta}={I}_{{n}} \:\:{we}\:{have}\:{to}\:{find}\:{I}_{\mathrm{10}} \\ $$$${I}_{{n}} =\mathrm{2}{I}_{{n}−\mathrm{1}} −{I}_{{n}−\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{I}_{\mathrm{0}} =\mathrm{0}\right] \\ $$$${I}_{\mathrm{2}} =\mathrm{2}{I}_{\mathrm{1}} −{I}_{\mathrm{0}} =\mathrm{2}{I}_{\mathrm{1}} −\mathrm{0}=\mathrm{2}×\frac{\pi}{\mathrm{2}}\:=\pi\:\:\:\:\:\left[{I}_{\mathrm{1}} =\frac{\pi}{\mathrm{2}}\right] \\ $$$${I}_{\mathrm{3}} =\mathrm{2}{I}_{\mathrm{2}} −{I}_{\mathrm{1}} =\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${I}_{\mathrm{4}} =\mathrm{2}\pi \\ $$$$….. \\ $$$${thus}\:{on}\:{calculation}\:\left({attached}\:{photo}\right)\:{we}\:{get} \\ $$$${I}_{\mathrm{10}} =\mathrm{5}\pi \\ $$$$\boldsymbol{{A}}\:{general}\:{formula}\:{thus}\:{obtained} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \left({n}\theta\right)}{{sin}^{\mathrm{2}} \theta}{d}\theta=\boldsymbol{{n}}\left(\frac{\pi}{\mathrm{2}}\right)={I}_{{n}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 29/Jan/19
$${Thank}\:{you}\:{sir}! \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19
![3)∫_0 ^(π/4) ((ln(cotx))/([(sinx)^(2009) +(cosx)^(2009) ]^2 )) (sin2x)^(2008) dx ∫_0 ^(π/4) ((ln(cotx))/([(sinx)^a +(cosx)^a ]^2 ))×(2sinxcosx)^(a−1) dx ∫_0 ^(π/4) ((ln(cotx))/((cosx)^(2a) [1+(tanx)^a ]^2 ))×(2tanxcos^2 x)^(a−1) dx ∫_0 ^(π/4) ((−ln(tanx))/((cosx)^(2a) [1+(tanx)^a ]^2 ))×(2)^(a−1) ×(tanx)^(a−1) ×(cosx)^(2a−2) dx =2^(a−1) ∫_0 ^(π/4) ((−ln(tanx))/([1+(tanx)^a ]^2 ))×sec^2 x×(tanx)^(a−1) dx let k=(tanx)^a (dk/dx)=a(tanx)^(a−1) ×sec^2 x (dk/a)=(tanx)^(a−1) ×sec^2 xdx lnk=a(lntanx) =2^(a−1) ∫_0 ^1 (((−lnk)/a)/([1+k]^2 ))×(dk/a) =−2^(a−1) ∫_0 ^1 ((lnk)/([1+k]^2 ))dk now ∫((lnk)/((1+k)^2 )) lnk∫(dk/((1+k)^2 ))−[(d/dk)(lnk)∫(dk/((1+k)^2 ))]dk =((−lnk)/((1+k)))−∫(1/k)×((−1)/(1+k))dk =((−lnk)/((1+k)))+∫((k+1−k)/(k(k+1)))dk =((−lnk)/((1+k)))+∫(dk/k)−∫(dk/(k+1)) =((−lnk)/((1+k)))+lnk−ln(k+1) =((−lnk)/((1+k)))+ln((k/(k+1)))+c so answer of (−2^(a−1) )∫_0 ^1 ((lnk)/((1+k)^2 ))dk is =(−2^(a−1) )∣((−lnk)/(1+k))+ln((k/(k+1)))∣_0 ^1 =(−2^(a−1) )[(((−ln1)/(1+1))+ln((1/2))+((ln(0))/(1+0))−ln(0)] so in think ln(0)−ln(0) cancelled and answer is =(−2^(a−1) )×(−ln2)=2^(a−1) ln2=2^(2008) ln2 is answer](https://www.tinkutara.com/question/Q54113.png)
$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left[\left({sinx}\right)^{\mathrm{2009}} +\left({cosx}\right)^{\mathrm{2009}} \right]^{\mathrm{2}} }\:\left({sin}\mathrm{2}{x}\right)^{\mathrm{2008}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left[\left({sinx}\right)^{{a}} +\left({cosx}\right)^{{a}} \right]^{\mathrm{2}} }×\left(\mathrm{2}{sinxcosx}\right)^{{a}−\mathrm{1}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left({cosx}\right)^{\mathrm{2}{a}} \left[\mathrm{1}+\left({tanx}\right)^{{a}} \right]^{\mathrm{2}} }×\left(\mathrm{2}{tanxcos}^{\mathrm{2}} {x}\right)^{{a}−\mathrm{1}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{−{ln}\left({tanx}\right)}{\left({cosx}\right)^{\mathrm{2}{a}} \left[\mathrm{1}+\left({tanx}\right)^{{a}} \right]^{\mathrm{2}} }×\left(\mathrm{2}\right)^{{a}−\mathrm{1}} ×\left({tanx}\right)^{{a}−\mathrm{1}} ×\left({cosx}\right)^{\mathrm{2}{a}−\mathrm{2}} {dx} \\ $$$$=\mathrm{2}^{{a}−\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{−{ln}\left({tanx}\right)}{\left[\mathrm{1}+\left({tanx}\right)^{{a}} \right]^{\mathrm{2}} }×{sec}^{\mathrm{2}} {x}×\left({tanx}\right)^{{a}−\mathrm{1}} {dx} \\ $$$${let}\:{k}=\left({tanx}\right)^{{a}} \:\:\:\:\frac{{dk}}{{dx}}={a}\left({tanx}\right)^{{a}−\mathrm{1}} ×{sec}^{\mathrm{2}} {x} \\ $$$$\frac{{dk}}{{a}}=\left({tanx}\right)^{{a}−\mathrm{1}} ×{sec}^{\mathrm{2}} {xdx} \\ $$$${lnk}={a}\left({lntanx}\right) \\ $$$$=\mathrm{2}^{{a}−\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{−{lnk}}{{a}}}{\left[\mathrm{1}+{k}\right]^{\mathrm{2}} }×\frac{{dk}}{{a}} \\ $$$$=−\mathrm{2}^{{a}−\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnk}}{\left[\mathrm{1}+{k}\right]^{\mathrm{2}} }{dk} \\ $$$${now}\:\int\frac{{lnk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} } \\ $$$${lnk}\int\frac{{dk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} }−\left[\frac{{d}}{{dk}}\left({lnk}\right)\int\frac{{dk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} }\right]{dk} \\ $$$$=\frac{−{lnk}}{\left(\mathrm{1}+{k}\right)}−\int\frac{\mathrm{1}}{{k}}×\frac{−\mathrm{1}}{\mathrm{1}+{k}}{dk} \\ $$$$=\frac{−{lnk}}{\left(\mathrm{1}+{k}\right)}+\int\frac{{k}+\mathrm{1}−{k}}{{k}\left({k}+\mathrm{1}\right)}{dk} \\ $$$$=\frac{−{lnk}}{\left(\mathrm{1}+{k}\right)}+\int\frac{{dk}}{{k}}−\int\frac{{dk}}{{k}+\mathrm{1}} \\ $$$$=\frac{−{lnk}}{\left(\mathrm{1}+{k}\right)}+{lnk}−{ln}\left({k}+\mathrm{1}\right) \\ $$$$=\frac{−{lnk}}{\left(\mathrm{1}+{k}\right)}+{ln}\left(\frac{{k}}{{k}+\mathrm{1}}\right)+{c} \\ $$$${so}\:{answer}\:{of}\:\left(−\mathrm{2}^{{a}−\mathrm{1}} \right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} }{dk}\:{is} \\ $$$$=\left(−\mathrm{2}^{{a}−\mathrm{1}} \right)\mid\frac{−{lnk}}{\mathrm{1}+{k}}+{ln}\left(\frac{{k}}{{k}+\mathrm{1}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$ \\ $$$$=\left(−\mathrm{2}^{{a}−\mathrm{1}} \right)\left[\left(\frac{−{ln}\mathrm{1}}{\mathrm{1}+\mathrm{1}}+{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{{ln}\left(\mathrm{0}\right)}{\mathrm{1}+\mathrm{0}}−{ln}\left(\mathrm{0}\right)\right]\right. \\ $$$${so}\:{in}\:{think}\:{ln}\left(\mathrm{0}\right)−{ln}\left(\mathrm{0}\right)\:{cancelled} \\ $$$${and}\:{answer}\:{is} \\ $$$$=\left(−\mathrm{2}^{{a}−\mathrm{1}} \right)×\left(−{ln}\mathrm{2}\right)=\mathrm{2}^{{a}−\mathrm{1}} {ln}\mathrm{2}=\mathrm{2}^{\mathrm{2008}} {ln}\mathrm{2}\:{is}\:{answer} \\ $$