Question Number 123859 by sdfg last updated on 28/Nov/20
$$\int{sect}\:{e}^{{t}} \:{dt} \\ $$
Answered by mindispower last updated on 29/Nov/20
![sec(x)=(2/(e^(ix) +e^(−ix) )),we use cos(x)=((e^(ix) +e^(−ix) )/2) =((2e^(ix) )/(1+e^(2ix) ))=2Σ_(n≥0) (−1)^n e^((2n+1)ix) ∫sec(x)e^x dx=2∫Σ_(n≥0) (−1)^n e^((2n+1)ix+x) dx 2Σ_(n≥0) (−1)^n ∫e^((2n+1)ix+x) dx =2Σ_(n≥0) (−1)^n (e^(((2n+1)i+1)x) /((2n+1)i+1))+c =2e^((1+i)x) Σ_(n≥0) (((−e^(2ix) )^n )/((2n+1)i+1)) =2e^((1+i)x) Σ_(n≥0) ((−i(−e^(2ix) )^n )/(2n+1−i))=e^((1+i)x) Σ_(n≥0) (((−i)(−e^(2ix) )^n )/(n+(((1−i)/2)))) =e^((1+i)x) Σ_(n≥0) (((−i)(−e^(2ix) )^n )/(n!)).((n!)/(n+(((1−i)/2)))) =e^((1+i)x) (((−2i)/(1−i))+Σ_(n≥1) (((−i)(−e^(2ix) )^n .Π_(k=0) ^(n−1) (k+(((1−i)/2)).Π_(k=0) ^(n−1) (1+k))/(Π_(k=0) ^(n−1) (k+(((3−i)/2))).(((1−i)/2)))) =−((2i)/(1−i))e^((1+i)x) (1+((1−i)/(−2i))Σ_(n≥1) (((−i))/((((1−i)/2)))).(((((1−i)/2))_n (1)_n )/((((3−i)/2))_n )).(((−e^(2ix) )^n )/(n!))) =(1−i)e^((1+i)x) (1+Σ_(n≥1) (((((1−i)/2))_n (1)_n )/((((3−i)/2))_n )).(((−e^(2ix) )^n )/(n!)))+c (a)_n =Π_(k=0) ^(n−1) (a+k),n!=1......n=(1+0)....(1+(n−1)) =(1)_n ....we used this all long now use 2F_1 (a,b;c;[x])=(1+Σ_(n≥1) (((a_n )(b)_n )/((c)_n )).(x^n /(n!))) we get =(1−i)e_ ^((1+i)x) _2 F_1 (((1−i)/2),1;((3−i)/2),[−e^(2ix) ])+C C constante ∫sec(x)e^x dx=(1−i)e^((1+i)x) _2 F_1 (((1−i)/2),1,((3−i)/2);[−e^(2ix) ])+C [X] is argument of function not floor function](Q123897.png)
$${sec}\left({x}\right)=\frac{\mathrm{2}}{{e}^{{ix}} +{e}^{−{ix}} },{we}\:{use}\:{cos}\left({x}\right)=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{e}^{{ix}} }{\mathrm{1}+{e}^{\mathrm{2}{ix}} }=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {e}^{\left(\mathrm{2}{n}+\mathrm{1}\right){ix}} \\ $$$$\int{sec}\left({x}\right){e}^{{x}} {dx}=\mathrm{2}\int\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {e}^{\left(\mathrm{2}{n}+\mathrm{1}\right){ix}+{x}} {dx} \\ $$$$\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int{e}^{\left(\mathrm{2}{n}+\mathrm{1}\right){ix}+{x}} {dx} \\ $$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{{e}^{\left(\left(\mathrm{2}{n}+\mathrm{1}\right){i}+\mathrm{1}\right){x}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){i}+\mathrm{1}}+{c} \\ $$$$=\mathrm{2}{e}^{\left(\mathrm{1}+{i}\right){x}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){i}+\mathrm{1}} \\ $$$$=\mathrm{2}{e}^{\left(\mathrm{1}+{i}\right){x}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{−{i}\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}−{i}}={e}^{\left(\mathrm{1}+{i}\right){x}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−{i}\right)\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{{n}+\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)} \\ $$$$={e}^{\left(\mathrm{1}+{i}\right){x}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−{i}\right)\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{{n}!}.\frac{{n}!}{{n}+\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)} \\ $$$$={e}^{\left(\mathrm{1}+{i}\right){x}} \left(\frac{−\mathrm{2}{i}}{\mathrm{1}−{i}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−{i}\right)\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} .\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+{k}\right)\right.}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\left(\frac{\mathrm{3}−{i}}{\mathrm{2}}\right)\right).\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)}\right. \\ $$$$=−\frac{\mathrm{2}{i}}{\mathrm{1}−{i}}{e}^{\left(\mathrm{1}+{i}\right){x}} \left(\mathrm{1}+\frac{\mathrm{1}−{i}}{−\mathrm{2}{i}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−{i}\right)}{\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)}.\frac{\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} }{\left(\frac{\mathrm{3}−{i}}{\mathrm{2}}\right)_{{n}} }.\frac{\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{{n}!}\right) \\ $$$$=\left(\mathrm{1}−{i}\right){e}^{\left(\mathrm{1}+{i}\right){x}} \left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} }{\left(\frac{\mathrm{3}−{i}}{\mathrm{2}}\right)_{{n}} }.\frac{\left(−{e}^{\mathrm{2}{ix}} \right)^{{n}} }{{n}!}\right)+{c} \\ $$$$\left({a}\right)_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}+{k}\right),{n}!=\mathrm{1}......{n}=\left(\mathrm{1}+\mathrm{0}\right)....\left(\mathrm{1}+\left({n}−\mathrm{1}\right)\right) \\ $$$$=\left(\mathrm{1}\right)_{{n}} ....{we}\:{used}\:{this}\:{all}\:\:{long} \\ $$$${now}\:{use}\:\mathrm{2}{F}_{\mathrm{1}} \left({a},{b};{c};\left[{x}\right]\right)=\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({a}_{{n}} \right)\left({b}\right)_{{n}} }{\left({c}\right)_{{n}} }.\frac{{x}^{{n}} }{{n}!}\right) \\ $$$${we}\:{get} \\ $$$$=\left(\mathrm{1}−{i}\right){e}_{\:} ^{\left(\mathrm{1}+{i}\right){x}} \:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}−{i}}{\mathrm{2}},\mathrm{1};\frac{\mathrm{3}−{i}}{\mathrm{2}},\left[−{e}^{\mathrm{2}{ix}} \right]\right)+{C} \\ $$$${C}\:{constante} \\ $$$$\int{sec}\left({x}\right){e}^{{x}} {dx}=\left(\mathrm{1}−{i}\right){e}^{\left(\mathrm{1}+{i}\right){x}} \:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}−{i}}{\mathrm{2}},\mathrm{1},\frac{\mathrm{3}−{i}}{\mathrm{2}};\left[−{e}^{\mathrm{2}{ix}} \right]\right)+{C} \\ $$$$\left[{X}\right]\:{is}\:{argument}\:{of}\:{function}\:{not}\:{floor}\:{function} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$