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14. Integration of the square-root of tan x 본문

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14. Integration of the square-root of tan x

Cambridge Maths Academy 2021. 11. 23. 22:37
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Consider I=tanxdxfor0x<π2

 

(a) Method 1. By using the substitution u=tanx, followed by partial fractions, we can show that I=122ln|tanx2tanx+1tanx+2tanx+1|+12arctan(tanx12tanx)+C=122ln|tanx+cotx2tanx+cotx+2|+12arctan(tanxcotx2)+C

 

(b) Method 2. By using the substitution u=tanx, followed by two separate substitutions z=u+1u and t=u1u, we can derive the same result as (a).

 

(c) Method 3. By considering I together with J=cotx, we can show that I=12arcsin(sinxcosx)12arcosh(sinx+cosx)+C=12arcsin[2sin(xπ4)]12arcosh[2sin(x+π4)]+C=12arcsin(sinxcosx)12ln(sinx+cosx+sin2x)+C

 

Solution.

 

(a) Method 1. Substitution u=tanx, followed by partial fractions.

 

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Step 1. Use the substitution, u=tanx=(tanx)12(u0)dudx=12(tanx)12sec2x=1+tan2x2tanx=1+u42u and the integral becomes: I=(tanxdxdu)du=(u×2u1+u4)du=2u21+u4du

 

Step 2. We try to factorise the denominator. For 1+u4, we can think of the following two possibilities. (1)(1+au+u2)(1+bu+u2)=1+(a+b)u+(ab+2)u2+(a+b)u3+u4(2)(1+auu2)(1+buu2)=1+(a+b)u+(ab2)u2(a+b)u3+u4 The linear and cubic terms require a+b=0. The quadratic term then reads (a2±2)u2 and, for aR, only the positive sign is possible, i.e. we choose (1) with a=b=2, i.e. 1+u4=(12u+u2)(1+2u+u2) The integrand then reads by partial fractions: 2u(12u+u2)(1+2u+u2)=A+Bu12u+u2+C+Du1+2u+u2=(A+Bu)(1+2u+u2)+(C+Du)(12u+u2)(12u+u2)(1+2u+u2)=(A+C)+[2(AC)+(B+D)]u+[(A+C)+2(BD)]u2+(B+D)u3(12u+u2)(1+2u+u2) The constant and cubic terms require A+C=0 and B+D=0. The linear term then sets A=C=0 and the quadratic term gives B=D=12.

 

Step 3. I=2u(12u+u2)(1+2u+u2)du=(12u12u+u212u1+2u+u2)du(by partial fractions)=122(2u12u+u22u1+2u+u2)du=122((2+2u)+212u+u2(2+2u)21+2u+u2)du=122(2+2u12u+u2+212u+u22+2u1+2u+u2+21+2u+u2)du=122(ln|12u+u2|ln|1+2u+u2|)+12(112u+u2+11+2u+u2)du=122ln|12u+u21+2u+u2|+12(1(u12)2+12+1(u+12)2+12)du=122ln|12u+u21+2u+u2|+122arctan[2(u12)]+122arctan[2(u+12)]+C=122ln|12u+u21+2u+u2|+12[arctan(2u1)+arctan(2u+1)]+C since 1(xa)2+b2dx=1barctan(xab)+C

 

Step 4. We can simplify the final expression slightly further by recalling: arctanx+arctany={arctan(x+y1xy)+π,xy>1,x>0,y>0(y>1x)arctan(x+y1xy)π,xy>1,x<0,y<0(y<1x)arctan(x+y1xy),otherwisearctan(2u1)+arctan(2u+1)=arctan[(2u1)+(2u+1)1(2u1)(2u+1)]=arctan(2u1u2)arctan(2u1)+arctan(2u+1)={arctan(2u1u2)+πforu>1arctan(2u1u2)πforu<1arctan(2u1u2)for1u1

 

which gives arctan(2u1)+arctan(2u+1)=arctan(u212u)+sgn(u)π2 where, for the last step, we used arctanx+arccotx=π2arctanx+arctan(1x)={π2,x>0π2,x<0 and tan(x)=tanx.

 

Finally, noting that u=tanx0, we find: I=122ln|12u+u21+2u+u2|+12arctan(u212u)+sgn(u)π22+C=C=122ln|12tanx+tanx1+2tanx+tanx|+12arctan(tanx12tanx)+C=122ln|tanx+cotx2tanx+cotx+2|+12arctan(tanxcotx2)+C

 


 

(b) Method 2. Substitution u=tanx, followed by two separate substitutions.

 

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Step 1. Use the substitution, u=tanx=(tanx)12(0x<π2,u0)dudx=12(tanx)12sec2x=1+tan2x2tanx=1+u42u and the integral becomes: I=(tanxdxdu)du=(u×2u1+u4)du=2u21+u4du=2u2+1u2du=[11u2+1+1u2u2+1u2]du=[11u2(u+1u)22+1+1u2(u1u)2+2]du

 

Step 2. Use the substitutions (for u0): z=u+1u(z2)andt=u1u(tR)dz=(11u2)duanddt=(1+1u2)du The integral becomes I=1z22dz+1t2+2dt=1(z2)(z+2)dz+12arctan(t2)+C=122(1z21z+2)dz+12arctan(t2)+C=122(ln|z2|ln|z+2|)+12arctan(t2)+C=122ln|z2z+2|+12arctan(t2)+C

 

since 1x2a2dx=1aarcoth(xa)+C=12aln(xax+a)+C(|x|>a)1a2x2dx=1aartanh(xa)+C=12aln(a+xax)+C(a<x<a)1x2+a2dx=1aarctan(xa)+C(xR) Finally, I=122ln|u+1u2u+1u+2|+12arctan(u1u2)+C=122ln|tanx+cotx2tanx+cotx+2|+12arctan(tanxcotx2)+C=122ln|tanx2tanx+1tanx+2tanx+1|+12arctan(tanx12tanx)+C

 


 

(c) Method 3. We consider the following two integrals: I=tanxdx,(0x<π2)J=cotxdx,(0<xπ2) Now, consider I+J and IJ.

 

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Step 1. For I+J, I+J=(tanx+cotx)dx=(sinxcosx+cosxsinx)dx=sinx+cosxsinxcosxdx=2sinx+cosx2sinxcosxdx We note that: (sinx+cosx)2=1+2sinxcosx(sinxcosx)2=12sinxcosx2sinxcosx=(sinx+cosx)21=1(sinxcosx)2 and that d(sinxcosx)=(cosx+sinx)dx, I+J can be written as: I+J=2sinx+cosx1(sinxcosx)2dx Use the substitution: u=sinxcosxdu=(cosx+sinx)dx and the integral becomes: I+J=211u2du=2arcsinu+C=2arcsin(sinxcosx)+C

 

Step 2. For IJ, IJ=(tanxcotx)dx=(sinxcosxcosxsinx)dx=sinxcosxsinxcosxdx=2sinxcosx2sinxcosxdx Recall: 2sinxcosx=(sinx+cosx)21=1(sinxcosx)2 and note that d(sinx+cosx)=(cosxsinx)dx, IJ can be written as: IJ=2sinx+cosx(sinx+cosx)21dx Use the substitution: u=sinx+cosxdu=(cosxsinx)dx and the integral becomes: IJ=21u21du=2arcoshu+C=2arcosh(sinx+cosx)+C=2ln(u+u21)+C=2ln(sinx+cosx+(sinx+cosx)21)+C=2ln(sinx+cosx+2sinxcosx)+C=2ln(sinx+cosx+sin2x)+C

 

Step 3. We collect the results: I+J=2arcsin(sinxcosx)+CIJ=2arcosh(sinx+cosx)+C=2ln(sinx+cosx+sin2x)+C and they yield: I=tanxdx=12arcsin(sinxcosx)12arcosh(sinx+cosx)+C=12arcsin[2sin(xπ4)]12arcosh[2sin(x+π4)]+C=12arcsin(sinxcosx)12ln(sinx+cosx+sin2x)+CJ=cotxdx=12arcsin(sinxcosx)+12arcosh(sinx+cosx)+C=12arcsin[2sin(xπ4)]+12arcosh[2sin(x+π4)]+C=12arcsin(sinxcosx)+12ln(sinx+cosx+sin2x)+C

 


 

Aside. Together with the results for J=cotxdx (see here), we can make a comparison:

 

I=tanxdx=12arcsin(sinxcosx)12arcosh(sinx+cosx)+C=12arcsin[2sin(xπ4)]12arcosh[2sin(x+π4)]+C=12arcsin(sinxcosx)12ln(sinx+cosx+sin2x)+C=122ln|tanx2tanx+1tanx+2tanx+1|+12arctan(tanx12tanx)+C=122ln|tanx+cotx2tanx+cotx+2|+12arctan(tanxcotx2)+C

 

J=cotxdx=12arcsin(sinxcosx)+12arcosh(sinx+cosx)+C=12arcsin[2sin(xπ4)]+12arcosh[2sin(x+π4)]+C=12arcsin(sinxcosx)+12ln(sinx+cosx+sin2x)+C=122ln|cotx2cotx+1cotx+2cotx+1|+12arccot(cotx12cotx)+C=122ln|cotx+tanx+2cotx+tanx2|+12arccot(cotxtanx2)+C

 

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