Verified Solution • 2025–26 Session moderate

QJEEM17Made1e: JEE Main 2017 Mathematics - Class 12 JEEM17Made1e Solution: Let If = , where C is a constant of integration, then the…

Let {I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right).

If I4+I6{I_4} + {I_6} = atan⁡5x+bx5+Ca{\tan ^5}x + b{x^5} + C , where C is a constant of integration,

then the ordered pair (a,b)\left( {a,b} \right) is equal to

Multiple Choice Options

  • A (15,0)\left( {{1 \over 5},0} \right)
  • B (15,−1)\left( {{1 \over 5}, - 1} \right)
  • C (−15,0)\left( { - {1 \over 5},0} \right)
  • D (−15,1)\left( { - {1 \over 5},1} \right)

Step-by-Step Verified Solution

Given,

In = \int {{{\tan }^n}x\,dx,\,\,\,n > 1}

∴   \therefore\,\,\, I4 = ∫tan⁡4x dx\int {{{\tan }^4}x\,dx}

and I6 = ∫tan⁡6x dx\int {{{\tan }^6}} x\,dx

∴   \therefore\,\,\, I = I4 + I6

= ∫(tan⁡4x+tan⁡6x)dx\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx

= ∫tan⁡4x(1+tan⁡2x)dx\int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx

= ∫tan⁡4x.sec⁡2x dx\int {{{\tan }^4}} x.{\sec ^2}x\,dx

Let, tanx = t

⇒\Rightarrow    \,\,\, sec2x dx = dt

∴   \therefore\,\,\, I = ∫t4 dt\int {{t^4}\,dt}

= 15{1 \over 5} t5 + C

= 15{1 \over 5} tan5x + C

∴   \therefore\,\,\, By comparing with the question, we get

A = 15{1 \over 5} ,  B = 0

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