Question Number 83108 by M±th+et£s last updated on 28/Feb/20
![prove that ∫_0 ^(π/4) ((cos(nx))/(cos^n (x))) dx =2^n [(π/8)−Σ_(k=1) ^(n−1) ((sin(((kπ)/4)))/(2k((√2))^k ))] n∈N^∗](https://www.tinkutara.com/question/Q83108.png)
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cos}\left({nx}\right)}{{cos}^{{n}} \left({x}\right)}\:{dx}\:=\mathrm{2}^{{n}} \left[\frac{\pi}{\mathrm{8}}−\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\mathrm{2}{k}\left(\sqrt{\mathrm{2}}\right)^{{k}} }\right]\:{n}\in{N}^{\ast} \\ $$
Answered by mind is power last updated on 28/Feb/20
![=Re∫_0 ^(π/4) (e^(inx) /((((e^(ix) +e^(−ix) )/2))^n )) =Re{∫_0 ^(π/4) ((2^n e^(i2nx) )/((e^(2ix) +1)^n ))dx} u=e^(2ix) du=2ie^(2ix) dx =Re{2^n ∫_C (u^(n−1) /((1+u)^n )).(du/(2i)) } C quart unite circle ∫_C^′ (z^(n−1) /((1+z)^n ))dz=0 withe C′=[0,1]∪[e^(it) , 0<t<(π/2)]∪[i,0] sice f(z)=(z^(n−1) /((1+z)^n ))(1/(2i)) is Holomorphic ∫_(C′) f(z)dz=0⇒∫_C f(z)=−∫_0 ^1 f(z)dz−∫_i ^0 f(z)dz Re{2^n ∫f(z)dz}=Re{−(2^n /(2i))∫_0 ^1 (z^(n−1) /((1+z)^n ))dz−(2^n /(2i))∫_i ^0 (z^(n−1) /((1+z)^n ))dz} =Re{(2^n /(2i))∫_0 ^i (z^(n−1) /((1+z)^n ))dz} (x^(n−1) /((1+x)^n ))=(((x+1−1)^(n−1) )/((1+x)^n )) =Σ_(k=0) ^(n−1) C_(n−1) ^k (x+1)^(n−1−k) (−1)^k =Σ_(k=0) ^(n−1) ((C_(n−1) ^k (−1)^k )/((x+1)^(1+k) ))==Σ_(j=1) ^n ((C_(n−1) ^(j−1) (−1)^(j−1) )/((x+1)^j )) ∫_0 ^i Σ_(j=1) ^n ((C_(n−1) ^(j−1) (−1)^(j−1) )/((x+1)^j ))dx =∫_0 ^i (dx/((x+1)))+Σ_(j=2) ^n ∫_0 ^i ((C_(n−1) ^(j−1) (−1)^(j−1) )/((x+1)^j )) =ln(1+i)+Σ_(j=2) ^n C_(n−1) ^(j−1) (−1)^(j−1) ∫_0 ^i (dx/((x+1)^j )) =ln((√2))+i(π/4)+Σ_(j=2) ^n C_(n−1) ^(j−1) (−1)^(j−1) .[(((x+1)^(1−j) )/(1−j))]_0 ^i =ln((√2))+i(π/4)+Σ_(j=2) ^n C_(n−1) ^(j−1) .(((1+i)^(1−j) −1)/(1−j)) =ln((√2))+((iπ)/4)+Σ_(j=2) ^n C_(n−1) ^(j−1) ((((√2))^(1−j) e^(i(((1−j)π)/4)) )/(1−j))−Σ_(j=2) ^n (C_(n−1) ^(j−1) /(1−j)) =((iπ)/4)+Σ_(j=1) ^(n−1) C_(n−1) ^j ((((√2))^(−j) e^(i(((jπ)/4))) )/(−j))−Σ_(j=1) ^(n−1) (C_(n−1) ^(−j) /(−j)) Re{(2^n /(2i)).[((iπ)/4)−Σ_(j=1) ^(n−1) C_(n−1) ^j (e^(i((jπ)/4)) /(j((√2))^j ))+Σ_(j=1) ^(n−1) (C_(n−1) ^j /j)]} =((2^n π)/8)−(1/2)Σ_(j=1) ^(n−1) C_(n−1) ^j ((2^n sin(((jπ)/4)))/(j((√2))^j )) =2^n ((π/8)−Σ_(k=1) ^(n−1) ((C_(n−1) ^k sin(((kπ)/4)))/(2k((√2))^k ))),check it Sir](https://www.tinkutara.com/question/Q83187.png)
$$={Re}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{e}^{{inx}} }{\left(\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right)^{{n}} }\: \\ $$$$={Re}\left\{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2}^{{n}} {e}^{{i}\mathrm{2}{nx}} }{\left({e}^{\mathrm{2}{ix}} +\mathrm{1}\right)^{{n}} }{dx}\right\} \\ $$$${u}={e}^{\mathrm{2}{ix}} \\ $$$${du}=\mathrm{2}{ie}^{\mathrm{2}{ix}} {dx} \\ $$$$={Re}\left\{\mathrm{2}^{{n}} \int_{{C}} \frac{{u}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{{n}} }.\frac{{du}}{\mathrm{2}{i}}\:\right\}\:\:\:{C}\:\:\:{quart}\:{unite}\:{circle} \\ $$$$\int_{{C}^{'} } \frac{{z}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{z}\right)^{{n}} }{dz}=\mathrm{0}\:\:\:{withe}\:{C}'=\left[\mathrm{0},\mathrm{1}\right]\cup\left[{e}^{{it}} ,\:\mathrm{0}<{t}<\frac{\pi}{\mathrm{2}}\right]\cup\left[{i},\mathrm{0}\right] \\ $$$${sice}\:{f}\left({z}\right)=\frac{{z}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{z}\right)^{{n}} }\frac{\mathrm{1}}{\mathrm{2}{i}}\:{is}\:{Holomorphic} \\ $$$$\int_{{C}'} {f}\left({z}\right){dz}=\mathrm{0}\Rightarrow\int_{{C}} {f}\left({z}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({z}\right){dz}−\int_{{i}} ^{\mathrm{0}} {f}\left({z}\right){dz} \\ $$$${Re}\left\{\mathrm{2}^{{n}} \int{f}\left({z}\right){dz}\right\}={Re}\left\{−\frac{\mathrm{2}^{{n}} }{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{z}\right)^{{n}} }{dz}−\frac{\mathrm{2}^{{n}} }{\mathrm{2}{i}}\int_{{i}} ^{\mathrm{0}} \frac{{z}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{z}\right)^{{n}} }{dz}\right\} \\ $$$$={Re}\left\{\frac{\mathrm{2}^{{n}} }{\mathrm{2}{i}}\int_{\mathrm{0}} ^{{i}} \frac{{z}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{z}\right)^{{n}} }{dz}\right\} \\ $$$$\frac{{x}^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{x}\right)^{{n}} }=\frac{\left({x}+\mathrm{1}−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{x}\right)^{{n}} } \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{k}} \left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}−{k}} \left(−\mathrm{1}\right)^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} }{\left({x}+\mathrm{1}\right)^{\mathrm{1}+{k}} }==\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \left(−\mathrm{1}\right)^{{j}−\mathrm{1}} }{\left({x}+\mathrm{1}\right)^{{j}} } \\ $$$$\int_{\mathrm{0}} ^{{i}} \underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \left(−\mathrm{1}\right)^{{j}−\mathrm{1}} }{\left({x}+\mathrm{1}\right)^{{j}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{{i}} \frac{{dx}}{\left({x}+\mathrm{1}\right)}+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\int_{\mathrm{0}} ^{{i}} \frac{{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \left(−\mathrm{1}\right)^{{j}−\mathrm{1}} }{\left({x}+\mathrm{1}\right)^{{j}} } \\ $$$$={ln}\left(\mathrm{1}+{i}\right)+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \left(−\mathrm{1}\right)^{{j}−\mathrm{1}} \int_{\mathrm{0}} ^{{i}} \frac{{dx}}{\left({x}+\mathrm{1}\right)^{{j}} } \\ $$$$={ln}\left(\sqrt{\mathrm{2}}\right)+{i}\frac{\pi}{\mathrm{4}}+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \left(−\mathrm{1}\right)^{{j}−\mathrm{1}} .\left[\frac{\left({x}+\mathrm{1}\right)^{\mathrm{1}−{j}} }{\mathrm{1}−{j}}\right]_{\mathrm{0}} ^{{i}} \\ $$$$={ln}\left(\sqrt{\mathrm{2}}\right)+{i}\frac{\pi}{\mathrm{4}}+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} .\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{1}−{j}} −\mathrm{1}}{\mathrm{1}−{j}} \\ $$$$={ln}\left(\sqrt{\mathrm{2}}\right)+\frac{{i}\pi}{\mathrm{4}}+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} \frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{1}−{j}} {e}^{{i}\frac{\left(\mathrm{1}−{j}\right)\pi}{\mathrm{4}}} }{\mathrm{1}−{j}}−\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{j}−\mathrm{1}} }{\mathrm{1}−{j}} \\ $$$$=\frac{{i}\pi}{\mathrm{4}}+\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}} \frac{\left(\sqrt{\mathrm{2}}\right)^{−{j}} {e}^{{i}\left(\frac{{j}\pi}{\mathrm{4}}\right)} }{−{j}}−\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{−{j}} }{−{j}} \\ $$$${Re}\left\{\frac{\mathrm{2}^{{n}} }{\mathrm{2}{i}}.\left[\frac{{i}\pi}{\mathrm{4}}−\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}} \frac{{e}^{{i}\frac{{j}\pi}{\mathrm{4}}} }{{j}\left(\sqrt{\mathrm{2}}\right)^{{j}} }+\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{j}} }{{j}}\right]\right\} \\ $$$$=\frac{\mathrm{2}^{{n}} \pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{j}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{n}−\mathrm{1}} ^{{j}} \frac{\mathrm{2}^{{n}} {sin}\left(\frac{{j}\pi}{\mathrm{4}}\right)}{{j}\left(\sqrt{\mathrm{2}}\right)^{{j}} } \\ $$$$=\mathrm{2}^{{n}} \left(\frac{\pi}{\mathrm{8}}−\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{C}_{{n}−\mathrm{1}} ^{{k}} {sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\mathrm{2}{k}\left(\sqrt{\mathrm{2}}\right)^{{k}} }\right),{check}\:{it}\:{Sir} \\ $$$$ \\ $$$$ \\ $$