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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2014 · Mathematics · Indefinite Integration

Evaluate: ∫ 1 1+3 ^ 2 x+8 ^ 2 x dx

Evaluate: ∫11+3sin⁡2x+8cos⁡2xdx\int \frac{1}{1+3 \sin ^{2} x+8 \cos ^{2} x} \mathrm{dx}

Options

  1. A16tan⁡−1(2tan⁡x)+C\frac{1}{6} \tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}
  2. Btan⁡−1(2tan⁡x)+C\tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}
  3. C\frac{1}{6} \tan ^{-1}(2tan⁡x3)+C\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}
  4. DNone of these

Correct answer

C. \frac{1}{6} \tan ^{-1}(2tan⁡x3)+C\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}

Step-by-step solution

I=∫11+3sin⁡2x+8cos⁡2xdx\mathrm{I}=\int \frac{1}{1+3 \sin ^{2} \mathrm{x}+8 \cos ^{2} \mathrm{x}} \mathrm{dx} Dividing the numerator and denominator by cos⁡2x,\cos ^{2} \mathrm{x}, we get I=∫sec⁡2xsec⁡2x+3tan⁡2x+8\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x+8} d x=∫sec⁡2x4tan⁡2x+9\int \frac{\sec ^{2} x}{4 \tan ^{2} x+9} d x Putting tan⁡x=t⇒sec⁡2xdx=dt,\tan x=t \Rightarrow \sec ^{2} x d x=d t, we get I=∫dt4\mathrmt2+9=14∫dt\mathrmt2+(3/2)2=14×13/2tan⁡−1\mathrm{I}=\int \frac{\mathrm{dt}}{4 \mathrmt^{2}+9}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrmt^{2}+(3 / 2)^{2}}=\frac{1}{4} \times \frac{1}{3 / 2} \tan ^{-1}\left(\frac{\mathrm{t}}{3 / 2}\right)+\mathrm{C}\Rightarrow I\mathrm{I}=\frac{1}{6} \tan ^{-1}(2t3)+C=16tan⁡−1(2tan⁡x3)+C\left(\frac{2 \mathrm{t}}{3}\right)+\mathrm{C}=\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}
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