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Indefinite Integrals question

2017 · Shift 0 · Q39
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Indefinite Integrals question

2017 · Shift 0 · Q39

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let In=∫tan⁡nx dx, (n>1).{I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n \gt 1} \right).In​=∫tannxdx,(n>1). If I4+I6{I_4} + {I_6}I4​+I6​=atan⁡5x+bx5+Ca{\tan ^5}x + b{x^5} + Catan5x+bx5+C, where C is a constant of integration, then the ordered pair (a,b)\left( {a,b} \right)(a,b) is equal to
  1. A
    (15,0)\left( {{1 \over 5},0} \right)(51​,0)
  2. B
    (15,−1)\left( {{1 \over 5}, - 1} \right)(51​,−1)
  3. C
    (−15,0)\left( { - {1 \over 5},0} \right)(−51​,0)
  4. D
    (−15,1)\left( { - {1 \over 5},1} \right)(−51​,1)
View written solutionFree

Correct answer: A

  1. We need to compute I4+I6=∫tan⁡4x dx+∫tan⁡6x dx=∫(tan⁡4x+tan⁡6x) dx.I_4+I_6=\int \tan^4 x\,dx+\int \tan^6 x\,dx=\int (\tan^4 x+\tan^6 x)\,dx.I4​+I6​=∫tan4xdx+∫tan6xdx=∫(tan4x+tan6x)dx.

  2. Factor the integrand: tan⁡4x+tan⁡6x=tan⁡4x(1+tan⁡2x).\tan^4 x+\tan^6 x=\tan^4 x(1+\tan^2 x).tan4x+tan6x=tan4x(1+tan2x). Using the identity 1+tan⁡2x=sec⁡2x,1+\tan^2 x=\sec^2 x,1+tan2x=sec2x, we get tan⁡4x+tan⁡6x=tan⁡4xsec⁡2x.\tan^4 x+\tan^6 x=\tan^4 x\sec^2 x.tan4x+tan6x=tan4xsec2x.

So, I4+I6=∫tan⁡4xsec⁡2x dx.I_4+I_6=\int \tan^4 x\sec^2 x\,dx.I4​+I6​=∫tan4xsec2xdx.

  1. Substitute t=tan⁡x  ⟹  dt=sec⁡2x dx.t=\tan x \implies dt=\sec^2 x\,dx.t=tanx⟹dt=sec2xdx. Then I4+I6=∫t4 dt=t55+C=tan⁡5x5+C.I_4+I_6=\int t^4\,dt=\frac{t^5}{5}+C=\frac{\tan^5 x}{5}+C.I4​+I6​=∫t4dt=5t5​+C=5tan5x​+C.

  2. Compare with the given form: I4+I6=atan⁡5x+bx5+C.I_4+I_6=a\tan^5 x+b x^5+C.I4​+I6​=atan5x+bx5+C. Hence, a=15,b=0.a=\frac{1}{5},\qquad b=0.a=51​,b=0.

  3. Therefore, the ordered pair is (a,b)=(15,0).\left(a,b\right)=\left(\frac{1}{5},0\right).(a,b)=(51​,0).

  4. Checking options:

  • A: (15,0)\left(\frac{1}{5},0\right)(51​,0) ✅
  • B, C, D are incorrect.

Thus the correct option is A.

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