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Question Number 123859 by sdfg last updated on 28/Nov/20

∫sect e^t dt

∫sectetdt

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

sec(x)=2eix+e−ix,weusecos(x)=eix+e−ix2=2eix1+e2ix=2∑n⩾0(−1)ne(2n+1)ix∫sec(x)exdx=2∫∑n⩾0(−1)ne(2n+1)ix+xdx2∑n⩾0(−1)n∫e(2n+1)ix+xdx=2∑n⩾0(−1)ne((2n+1)i+1)x(2n+1)i+1+c=2e(1+i)x∑n⩾0(−e2ix)n(2n+1)i+1=2e(1+i)x∑n⩾0−i(−e2ix)n2n+1−i=e(1+i)x∑n⩾0(−i)(−e2ix)nn+(1−i2)=e(1+i)x∑n⩾0(−i)(−e2ix)nn!.n!n+(1−i2)=e(1+i)x(−2i1−i+∑n⩾1(−i)(−e2ix)n.∏n−1k=0(k+(1−i2).∏n−1k=0(1+k)∏n−1k=0(k+(3−i2)).(1−i2)=−2i1−ie(1+i)x(1+1−i−2i∑n⩾1(−i)(1−i2).(1−i2)n(1)n(3−i2)n.(−e2ix)nn!)=(1−i)e(1+i)x(1+∑n⩾1(1−i2)n(1)n(3−i2)n.(−e2ix)nn!)+c(a)n=∏n−1k=0(a+k),n!=1......n=(1+0)....(1+(n−1))=(1)n....weusedthisalllongnowuse2F1(a,b;c;[x])=(1+∑n⩾1(an)(b)n(c)n.xnn!)weget=(1−i)e(1+i)x2F1(1−i2,1;3−i2,[−e2ix])+CCconstante∫sec(x)exdx=(1−i)e(1+i)x2F1(1−i2,1,3−i2;[−e2ix])+C[X]isargumentoffunctionnotfloorfunction

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