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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2016 · 9 April · Online · Mathematics · Indefinite Integration

If ∫ dx ^ 3 x 2 2x = ( x) ^ A + C ( x) ^ B + k , where k is a...

If ∫dxcos⁡3x2sin⁡2x=(tan⁡x)A+C(tan⁡x)B+k∫ \dfrac{dx}{\cos^{3}x\sqrt{2\sin 2x}} = {\left(\tan x\right)}^{A} + C{\left(\tan x\right)}^{B} + k, where kk is a constant of integration, then A+B+CA + B + C equals

Options

  1. A165\dfrac{16}{5}
  2. B2710\dfrac{27}{10}
  3. C710\dfrac{7}{10}
  4. D215\dfrac{21}{5}

Correct answer

A. 165\dfrac{16}{5}

Step-by-step solution

∫dxcos⁡3x4sin⁡xcos⁡x=∫dx2cos⁡4xtan⁡x∫ \dfrac{dx}{\cos^{3}x\sqrt{4\sin x \cos x}} = ∫ \dfrac{dx}{2\cos^{4}x\sqrt{\tan x}}

Let, tan⁡x=t2⇒sec⁡2x=1+t4\tan x = {t}^{2} ⇒ \sec^{2}x = 1 + {t}^{4}

And sec⁡2xdx=2tdt\sec^{2}xdx = 2tdt

∴I=∫sec⁡4xdx2tan⁡x=∫sec⁡2x.sec⁡2xdx2tan⁡x∴ I = ∫ \dfrac{\sec^{4}xdx}{2\sqrt{\tan x}} = ∫ \dfrac{\sec^{2}x . \sec^{2}xdx}{2\sqrt{\tan x}}

=∫(1+t4)2tdt2t=∫(1+t4)dt= ∫ \dfrac{\left(1 + {t}^{4}\right)2tdt}{2t} = ∫ \left(1 + {t}^{4}\right)dt

=t+t55+k= t + \dfrac{{t}^{5}}{5} + k, where kk is a constant of integration.

=tan⁡x+15tan52x+k= \sqrt{\tan x} + \dfrac{1}{5}tan\dfrac{5}{2}x + k

Where, t=tan⁡xt = \sqrt{\tan x}.

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

A+B+C=165A + B + C = \dfrac{16}{5}

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