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Cambridge Maths Academy
14. Integration of the square-root of tan x 본문
14. Integration of the square-root of tan x
Cambridge Maths Academy 2021. 11. 23. 22:37
수학 모음 (Maths collection) 전체보기
Consider I=∫√tanxdxfor0≤x<π2(a) Method 1. By using the substitution u=√tanx, followed by partial fractions, we can show that I=12√2ln|tanx−√2tanx+1tanx+√2tanx+1|+1√2arctan(tanx−1√2tanx)+C=12√2ln|√tanx+√cotx−√2√tanx+√cotx+√2|+1√2arctan(√tanx−√cotx√2)+C
(b) Method 2. By using the substitution u=√tanx, followed by two separate substitutions z=u+1u and t=u−1u, we can derive the same result as (a).
(c) Method 3. By considering I together with J=√cotx, we can show that I=1√2arcsin(sinx−cosx)−1√2arcosh(sinx+cosx)+C=1√2arcsin[√2sin(x−π4)]−1√2arcosh[√2sin(x+π4)]+C=1√2arcsin(sinx−cosx)−1√2ln(sinx+cosx+√sin2x)+C
Solution.
(a) Method 1. Substitution u=√tanx, followed by partial fractions.
Step 1. Use the substitution, u=√tanx=(tanx)12(u≥0)⇒dudx=12(tanx)−12sec2x=1+tan2x2√tanx=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+au−u2)(1+bu−u2)=1+(a+b)u+(ab−2)u2−(a+b)u3+u4 The linear and cubic terms require a+b=0. The quadratic term then reads (−a2±2)u2 and, for a∈R, only the positive sign is possible, i.e. we choose (1) with a=−b=√2, i.e. 1+u4=(1−√2u+u2)(1+√2u+u2) The integrand then reads by partial fractions: 2u(1−√2u+u2)(1+√2u+u2)=A+Bu1−√2u+u2+C+Du1+√2u+u2=(A+Bu)(1+√2u+u2)+(C+Du)(1−√2u+u2)(1−√2u+u2)(1+√2u+u2)=(A+C)+[√2(A−C)+(B+D)]u+[(A+C)+√2(B−D)]u2+(B+D)u3(1−√2u+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=1√2.
Step 3. ⇒I=∫2u(1−√2u+u2)(1+√2u+u2)du=∫(1√2u1−√2u+u2−1√2u1+√2u+u2)du(by partial fractions)=∫12√2(2u1−√2u+u2−2u1+√2u+u2)du=∫12√2((−√2+2u)+√21−√2u+u2−(√2+2u)−√21+√2u+u2)du=∫12√2(−√2+2u1−√2u+u2+√21−√2u+u2−√2+2u1+√2u+u2+√21+√2u+u2)du=12√2(ln|1−√2u+u2|−ln|1+√2u+u2|)+∫12(11−√2u+u2+11+√2u+u2)du=12√2ln|1−√2u+u21+√2u+u2|+12∫(1(u−1√2)2+12+1(u+1√2)2+12)du=12√2ln|1−√2u+u21+√2u+u2|+12⋅√2arctan[√2(u−1√2)]+12⋅√2arctan[√2(u+1√2)]+C′=12√2ln|1−√2u+u21+√2u+u2|+1√2[arctan(√2u−1)+arctan(√2u+1)]+C′ since ∫1(x−a)2+b2dx=1barctan(x−ab)+C
Step 4. We can simplify the final expression slightly further by recalling: arctanx+arctany={arctan(x+y1−xy)+π,xy>1,x>0,y>0(y>1x)arctan(x+y1−xy)−π,xy>1,x<0,y<0(y<1x)arctan(x+y1−xy),otherwise⇒arctan(√2u−1)+arctan(√2u+1)=arctan[(√2u−1)+(√2u+1)1−(√2u−1)(√2u+1)]=arctan(√2u1−u2)⇒arctan(√2u−1)+arctan(√2u+1)={arctan(√2u1−u2)+πforu>1arctan(√2u1−u2)−πforu<−1arctan(√2u1−u2)for−1≤u≤1
which gives arctan(√2u−1)+arctan(√2u+1)=arctan(u2−1√2u)+sgn(u)π2 where, for the last step, we used arctanx+arccotx=π2⇔arctanx+arctan(1x)={π2,x>0−π2,x<0 and tan(−x)=−tanx.
Finally, noting that u=√tanx≥0, we find: ⇒I=12√2ln|1−√2u+u21+√2u+u2|+1√2arctan(u2−1√2u)+sgn(u)π2√2+C′⏟=C=12√2ln|1−√2tanx+tanx1+√2tanx+tanx|+1√2arctan(tanx−1√2tanx)+C=12√2ln|√tanx+√cotx−√2√tanx+√cotx+√2|+1√2arctan(√tanx−√cotx√2)+C◻
(b) Method 2. Substitution u=√tanx, followed by two separate substitutions.
Step 1. Use the substitution, u=√tanx=(tanx)12(0≤x<π2,u≥0)⇒dudx=12(tanx)−12sec2x=1+tan2x2√tanx=1+u42u and the integral becomes: I=∫(√tanxdxdu)du=∫(u×2u1+u4)du=∫2u21+u4du=∫2u2+1u2du=∫[1−1u2+1+1u2u2+1u2]du=∫[1−1u2(u+1u)2−2+1+1u2(u−1u)2+2]du
Step 2. Use the substitutions (for u≥0): z=u+1u(z≥2)andt=u−1u(t∈R)dz=(1−1u2)duanddt=(1+1u2)du The integral becomes ⇒I=∫1z2−2dz+∫1t2+2dt=∫1(z−√2)(z+√2)dz+1√2arctan(t√2)+C=∫12√2(1z−√2−1z+√2)dz+1√2arctan(t√2)+C=12√2(ln|z−√2|−ln|z+√2|)+1√2arctan(t√2)+C′=12√2ln|z−√2z+√2|+1√2arctan(t√2)+C′
since ∫1x2−a2dx=−1aarcoth(xa)+C=12aln(x−ax+a)+C(|x|>a)∫1a2−x2dx=1aartanh(xa)+C=12aln(a+xa−x)+C(−a<x<a)∫1x2+a2dx=1aarctan(xa)+C(x∈R) Finally, I=12√2ln|u+1u−√2u+1u+√2|+1√2arctan(u−1u√2)+C=12√2ln|√tanx+√cotx−√2√tanx+√cotx+√2|+1√2arctan(√tanx−√cotx√2)+C=12√2ln|tanx−√2tanx+1tanx+√2tanx+1|+1√2arctan(tanx−1√2tanx)+C◻
(c) Method 3. We consider the following two integrals: I=∫√tanxdx,(0≤x<π2)J=∫√cotxdx,(0<x≤π2) Now, consider I+J and I−J.
Step 1. For I+J, I+J=∫(√tanx+√cotx)dx=∫(√sinx√cosx+√cosx√sinx)dx=∫sinx+cosx√sinxcosxdx=√2∫sinx+cosx√2sinxcosxdx We note that: (sinx+cosx)2=1+2sinxcosx(sinx−cosx)2=1−2sinxcosx⇒2sinxcosx=(sinx+cosx)2−1=1−(sinx−cosx)2 and that d(sinx−cosx)=(cosx+sinx)dx, I+J can be written as: ⇒I+J=√2∫sinx+cosx√1−(sinx−cosx)2dx Use the substitution: u=sinx−cosxdu=(cosx+sinx)dx and the integral becomes: ⇒I+J=√2∫1√1−u2du=√2arcsinu+C′=√2arcsin(sinx−cosx)+C′
Step 2. For I−J, I−J=∫(√tanx−√cotx)dx=∫(√sinx√cosx−√cosx√sinx)dx=∫sinx−cosx√sinxcosxdx=√2∫sinx−cosx√2sinxcosxdx Recall: 2sinxcosx=(sinx+cosx)2−1=1−(sinx−cosx)2 and note that d(sinx+cosx)=(cosx−sinx)dx, I−J can be written as: ⇒I−J=√2∫sinx+cosx√(sinx+cosx)2−1dx Use the substitution: u=sinx+cosxdu=(cosx−sinx)dx and the integral becomes: ⇒I−J=−√2∫1√u2−1du=−√2arcoshu+C″=−√2arcosh(sinx+cosx)+C″=−√2ln(u+√u2−1)+C″=−√2ln(sinx+cosx+√(sinx+cosx)2−1)+C″=−√2ln(sinx+cosx+√2sinxcosx)+C″=−√2ln(sinx+cosx+√sin2x)+C″
Step 3. We collect the results: I+J=√2arcsin(sinx−cosx)+C′I−J=−√2arcosh(sinx+cosx)+C″=−√2ln(sinx+cosx+√sin2x)+C″ and they yield: I=∫√tanxdx=1√2arcsin(sinx−cosx)−1√2arcosh(sinx+cosx)+C=1√2arcsin[√2sin(x−π4)]−1√2arcosh[√2sin(x+π4)]+C=1√2arcsin(sinx−cosx)−1√2ln(sinx+cosx+√sin2x)+CJ=∫√cotxdx=1√2arcsin(sinx−cosx)+1√2arcosh(sinx+cosx)+C=1√2arcsin[√2sin(x−π4)]+1√2arcosh[√2sin(x+π4)]+C=1√2arcsin(sinx−cosx)+1√2ln(sinx+cosx+√sin2x)+C
Aside. Together with the results for J=∫√cotxdx (see here), we can make a comparison:
I=∫√tanxdx=1√2arcsin(sinx−cosx)−1√2arcosh(sinx+cosx)+C=1√2arcsin[√2sin(x−π4)]−1√2arcosh[√2sin(x+π4)]+C=1√2arcsin(sinx−cosx)−1√2ln(sinx+cosx+√sin2x)+C=12√2ln|tanx−√2tanx+1tanx+√2tanx+1|+1√2arctan(tanx−1√2tanx)+C=12√2ln|√tanx+√cotx−√2√tanx+√cotx+√2|+1√2arctan(√tanx−√cotx√2)+C
J=∫√cotxdx=1√2arcsin(sinx−cosx)+1√2arcosh(sinx+cosx)+C=1√2arcsin[√2sin(x−π4)]+1√2arcosh[√2sin(x+π4)]+C=1√2arcsin(sinx−cosx)+1√2ln(sinx+cosx+√sin2x)+C=12√2ln|cotx−√2cotx+1cotx+√2cotx+1|+1√2arccot(cotx−1√2cotx)+C=12√2ln|√cotx+√tanx+√2√cotx+√tanx−√2|+1√2arccot(√cotx−√tanx√2)+C
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