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QJEEM16Ma7e73: JEE Main 2016 Mathematics - Class 12 JEEM16Ma7e73 Solution: If    where k is a constant of integration, then A +…

If    ∫dxcos⁡3x2sin⁡2x=(tan⁡x)A+C(tan⁡x)B+k,\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,

where k is a constant of integration, then A + B +C equals :

Multiple Choice Options

  • A 215{{21} \over 5}
  • B 165{{16} \over 5}
  • C 710{{7} \over 10}
  • D 2710{{27} \over 10}

Step-by-Step Verified Solution

∫dxcos⁡3x2sin⁡2x\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}}

=   ∫dxcos⁡3x4sin⁡xcos⁡x\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}}

=   ∫dx2cos⁡4xtan⁡x\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}}

Let tan x   =   t2

⇒\Rightarrow    \,\,\, sec2xdx = 2t dt

as   sec2x = 1 + tan2x = 1 + t4

=   ∫sec⁡4x dx2tan⁡x\int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}}

=   ∫sec⁡2x(sec⁡2x dx)2tan⁡x\int {{{{{\sec }^2}x\left( {{{\sec }^2}x\,dx} \right)} \over {2\sqrt {\tan x} }}}

=   ∫(1+t4)2t dt2t\int {{{\left( {1 + {t^4}} \right)2t\,dt} \over {2t}}}

=   ∫(1+t4) dt\int {\left( {1 + {t^4}} \right)} \,dt

=  t + t55{{{t^5}} \over 5} + k

=    tan⁡x\sqrt {\tan x} + 15{1 \over 5} tan 52^{{5 \over 2}} x + k

By comparing with the given equation, we get

A = 12{1 \over 2} , B = 52{5 \over 2} , C = 15{1 \over 5}

∴   \therefore\,\,\, A + B + C = 165{{16} \over 5}

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