Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2023 · 29 Jan · Shift 2 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2023 · 29 Jan · Shift 2 · Q34

Definite Integration question

2023 · 29 Jan · Shift 2 · Q34

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of the integral ∫12(t4+1t6+1)dt\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt}∫12​(t6+1t4+1​)dt is
  1. A
    tan⁡−112−13tan⁡−18+π3{\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}tan−121​−31​tan−18+3π​
  2. B
    tan⁡−12−13tan⁡−18+π3{\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}tan−12−31​tan−18+3π​
  3. C
    tan⁡−12+13tan⁡−18−π3{\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}tan−12+31​tan−18−3π​
  4. D
    tan⁡−112+13tan⁡−18−π3{\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}tan−121​+31​tan−18−3π​
View written solutionFree

Correct answer: C

  1. We need to evaluate I=∫12t4+1t6+1 dt.I=\int_1^2 \frac{t^4+1}{t^6+1}\,dt.I=∫12​t6+1t4+1​dt.

  2. Factor the denominator: t6+1=(t2+1)(t4−t2+1).t^6+1=(t^2+1)(t^4-t^2+1).t6+1=(t2+1)(t4−t2+1). So, t4+1t6+1=t4+1(t2+1)(t4−t2+1).\frac{t^4+1}{t^6+1}=\frac{t^4+1}{(t^2+1)(t^4-t^2+1)}.t6+1t4+1​=(t2+1)(t4−t2+1)t4+1​.

  3. Use partial fractions in the form t4+1(t2+1)(t4−t2+1)=At2+1+Bt2+Ct4−t2+1.\frac{t^4+1}{(t^2+1)(t^4-t^2+1)}=\frac{A}{t^2+1}+\frac{Bt^2+C}{t^4-t^2+1}.(t2+1)(t4−t2+1)t4+1​=t2+1A​+t4−t2+1Bt2+C​. Multiplying through by (t2+1)(t4−t2+1)(t^2+1)(t^4-t^2+1)(t2+1)(t4−t2+1), t4+1=A(t4−t2+1)+(Bt2+C)(t2+1).t^4+1=A(t^4-t^2+1)+(Bt^2+C)(t^2+1).t4+1=A(t4−t2+1)+(Bt2+C)(t2+1). Expanding: t4+1=At4−At2+A+Bt4+Bt2+Ct2+C.t^4+1=A t^4-A t^2+A+B t^4+B t^2+C t^2+C.t4+1=At4−At2+A+Bt4+Bt2+Ct2+C. t4+1=(A+B)t4+(−A+B+C)t2+(A+C).t^4+1=(A+B)t^4+(-A+B+C)t^2+(A+C).t4+1=(A+B)t4+(−A+B+C)t2+(A+C). Comparing coefficients: [ A+B=1,\qquad -A+B+C=0,\qquad A+C=1. ] Solving gives A=23,B=13,C=13.A=\frac23,\qquad B=\frac13,\qquad C=\frac13.A=32​,B=31​,C=31​. Thus, t4+1t6+1=23(t2+1)+13⋅t2+1t4−t2+1.\frac{t^4+1}{t^6+1}=\frac{2}{3(t^2+1)}+\frac{1}{3}\cdot\frac{t^2+1}{t^4-t^2+1}.t6+1t4+1​=3(t2+1)2​+31​⋅t4−t2+1t2+1​.

  4. Hence I=23∫12dtt2+1+13∫12t2+1t4−t2+1 dt.I=\frac23\int_1^2 \frac{dt}{t^2+1}+\frac13\int_1^2 \frac{t^2+1}{t^4-t^2+1}\,dt.I=32​∫12​t2+1dt​+31​∫12​t4−t2+1t2+1​dt.

Now observe that t4−t2+1=t6+1t2+1,t^4-t^2+1=\frac{t^6+1}{t^2+1},t4−t2+1=t2+1t6+1​, but more usefully, t4−t2+1=(t2−12)2+34.t^4-t^2+1=(t^2-\tfrac12)^2+\tfrac34.t4−t2+1=(t2−21​)2+43​. A standard substitution works better: let u=t−1t.u=t-\frac1t.u=t−t1​. Then du=(1+1t2)dt=t2+1t2dt,du=\left(1+\frac1{t^2}\right)dt=\frac{t^2+1}{t^2}dt,du=(1+t21​)dt=t2t2+1​dt, so (t2+1)dt=t2 du.(t^2+1)dt=t^2\,du.(t2+1)dt=t2du. Also, u2=t2−2+1t2  ⟹  t2+1t2=u2+2.u^2=t^2-2+\frac1{t^2}\implies t^2+\frac1{t^2}=u^2+2.u2=t2−2+t21​⟹t2+t21​=u2+2. Now t4−t2+1=t2(t2−1+1t2)=t2(u2+1).t^4-t^2+1=t^2\left(t^2-1+\frac1{t^2}\right)=t^2(u^2+1).t4−t2+1=t2(t2−1+t21​)=t2(u2+1). Therefore, t2+1t4−t2+1dt=t2 dut2(u2+1)=duu2+1.\frac{t^2+1}{t^4-t^2+1}dt=\frac{t^2\,du}{t^2(u^2+1)}=\frac{du}{u^2+1}.t4−t2+1t2+1​dt=t2(u2+1)t2du​=u2+1du​.

So the second integral becomes ∫12t2+1t4−t2+1 dt=∫u(1)u(2)duu2+1.\int_1^2 \frac{t^2+1}{t^4-t^2+1}\,dt=\int_{u(1)}^{u(2)} \frac{du}{u^2+1}.∫12​t4−t2+1t2+1​dt=∫u(1)u(2)​u2+1du​. At the limits: u(1)=1−1=0,u(2)=2−12=32.u(1)=1-1=0,\qquad u(2)=2-\frac12=\frac32.u(1)=1−1=0,u(2)=2−21​=23​. Hence, ∫12t2+1t4−t2+1 dt=tan⁡−1(32).\int_1^2 \frac{t^2+1}{t^4-t^2+1}\,dt=\tan^{-1}\left(\frac32\right).∫12​t4−t2+1t2+1​dt=tan−1(23​).

Thus, I=23[tan⁡−1t]12+13tan⁡−1(32).I=\frac23\left[\tan^{-1}t\right]_1^2+\frac13\tan^{-1}\left(\frac32\right).I=32​[tan−1t]12​+31​tan−1(23​). So, I=23(tan⁡−12−π4)+13tan⁡−1(32).I=\frac23\left(\tan^{-1}2-\frac\pi4\right)+\frac13\tan^{-1}\left(\frac32\right).I=32​(tan−12−4π​)+31​tan−1(23​).

  1. Now simplify to match the options. We use the identity tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab)\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)tan−1a+tan−1b=tan−1(1−aba+b​) with quadrant adjustment.

First, tan⁡−12+tan⁡−1(32)=π−tan⁡−1(2+3/22⋅3/2−1)=π−tan⁡−1(7/22),\tan^{-1}2+\tan^{-1}\left(\frac32\right)=\pi-\tan^{-1}\left(\frac{2+3/2}{2\cdot 3/2-1}\right)=\pi-\tan^{-1}\left(\frac{7/2}{2}\right),tan−12+tan−1(23​)=π−tan−1(2⋅3/2−12+3/2​)=π−tan−1(27/2​), though this is not immediately convenient. A better identity is: 3tan⁡−12=π+tan⁡−1(2(3−4)1−12)=π+tan⁡−1(811),3\tan^{-1}2=\pi+\tan^{-1}\left(\frac{2(3-4)}{1-12}\right)=\pi+\tan^{-1}\left(\frac{8}{11}\right),3tan−12=π+tan−1(1−122(3−4)​)=π+tan−1(118​), which is also not ideal directly.

Instead, use tan⁡(3tan⁡−12)=3(2)−231−3(22)=6−81−12=−2−11=211.\tan\big(3\tan^{-1}2\big)=\frac{3(2)-2^3}{1-3(2^2)}=\frac{6-8}{1-12}=\frac{-2}{-11}=\frac{2}{11}.tan(3tan−12)=1−3(22)3(2)−23​=1−126−8​=−11−2​=112​. This route is messy.

Let us instead verify option C by differentiation/algebraic equivalence using tangent addition. From the options, we want to show I=tan⁡−12+13tan⁡−18−π3.I=\tan^{-1}2+\frac13\tan^{-1}8-\frac\pi3.I=tan−12+31​tan−18−3π​. So compare with our form: I=23tan⁡−12−π6+13tan⁡−1(32).I=\frac23\tan^{-1}2-\frac\pi6+\frac13\tan^{-1}\left(\frac32\right).I=32​tan−12−6π​+31​tan−1(23​). Multiply by 333: 3I=2tan⁡−12−π2+tan⁡−1(32).3I=2\tan^{-1}2-\frac\pi2+\tan^{-1}\left(\frac32\right).3I=2tan−12−2π​+tan−1(23​). We need this to equal 3tan⁡−12+tan⁡−18−π.3\tan^{-1}2+\tan^{-1}8-\pi.3tan−12+tan−18−π. That is equivalent to tan⁡−1(32)+π2=tan⁡−12+tan⁡−18.\tan^{-1}\left(\frac32\right)+\frac\pi2=\tan^{-1}2+\tan^{-1}8.tan−1(23​)+2π​=tan−12+tan−18. Now, tan⁡(tan⁡−12+tan⁡−18)=2+81−16=10−15=−23.\tan\left(\tan^{-1}2+\tan^{-1}8\right)=\frac{2+8}{1-16}=\frac{10}{-15}=-\frac23.tan(tan−12+tan−18)=1−162+8​=−1510​=−32​. Since both angles are acute, their sum lies in quadrant II, so tan⁡−12+tan⁡−18=π−tan⁡−1(23).\tan^{-1}2+\tan^{-1}8=\pi-\tan^{-1}\left(\frac23\right).tan−12+tan−18=π−tan−1(32​). Also, tan⁡−1(32)+tan⁡−1(23)=π2\tan^{-1}\left(\frac32\right)+\tan^{-1}\left(\frac23\right)=\frac\pi2tan−1(23​)+tan−1(32​)=2π​ because their product is 111. Thus, tan⁡−1(32)=π2−tan⁡−1(23).\tan^{-1}\left(\frac32\right)=\frac\pi2-\tan^{-1}\left(\frac23\right).tan−1(23​)=2π​−tan−1(32​). Hence, tan⁡−1(32)+π2=π−tan⁡−1(23)=tan⁡−12+tan⁡−18.\tan^{-1}\left(\frac32\right)+\frac\pi2=\pi-\tan^{-1}\left(\frac23\right)=\tan^{-1}2+\tan^{-1}8.tan−1(23​)+2π​=π−tan−1(32​)=tan−12+tan−18. So the two expressions are equal.

Therefore, I=tan⁡−12+13tan⁡−18−π3.I=\tan^{-1}2+\frac13\tan^{-1}8-\frac\pi3.I=tan−12+31​tan−18−3π​.

  1. Hence the correct option is C.\boxed{\text{C}}.C​.
PreviousNext

More from Definite Integration

  • The value of the integral 1/2∫2​xtan−1x​dx is equal to :2023 · MCQ
  • If [t] denotes the greatest integer ≤t, then the value of e3(e−1)​1∫2​x2e[x]+[x3]dx is :2023 · MCQ
  • limx→0​x448​∫0x​t6+1t3​ dt is equal to ​.2023 · Numerical
  • Let α∈(0,1) and β=loge​(1−α). Let Pn​(x)=x+2x2​+3x3​+...+nxn​,x∈(0,1). Then the integral 0∫α​1−tt50​dt…2023 · MCQ
  • The value of ∫3π​2π​​sinx(1+cosx)(2+3sinx)​dx is equal to :2023 · MCQ
  • Let α>0. If 0∫α​x+α​−x​x​ dx=1516+202​​, then α is equal to :2023 · MCQ
  • If ϕ(x)=x​1​4π​∫x​(42​sint−3ϕ′(t))dt,x>0, then ∅′(4π​) is equal to :2023 · MCQ
  • Let f(θ)=sinθ+−π/2∫π/2​(sinθ+tcosθ)f(t)dt. Then the value of ​∫0π/2​f(θ)dθ​ is ​.2022 · Numerical