Q71 / 125 · BITSAT 2014 · Mathematics

Evaluate: $\int \frac{1}{1+3 \sin ^{2} x+8 \cos ^{2} x} \mathrm{dx}$

  • A
    $\frac{1}{6} \tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}$
  • B
    $\tan ^{-1}(2 \tan \mathrm{x})+\mathrm{C}$
  • Correct answer
    C
    $\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}$
  • D
    None of these

Solution

$\mathrm{I}=\int \frac{1}{1+3 \sin ^{2} \mathrm{x}+8 \cos ^{2} \mathrm{x}} \mathrm{dx}$ Dividing the numerator and denominator by $\cos ^{2} \mathrm{x},$ we get $$ I=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x+8} d x=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+9} d x $$ Putting $\tan x=t \Rightarrow \sec ^{2} x d x=d t,$ we get $\mathrm{I}=\int \frac{\mathrm{dt}}{4 \mathrm{t}^{2}+9}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+(3 / 2)^{2}}=\frac{1}{4} \times \frac{1}{3 / 2} \tan ^{-1}\left(\frac{\mathrm{t}}{3 / 2}\right)+\mathrm{C}$ $\Rightarrow \mathrm{I}=\frac{1}{6} \tan ^{-1}\left(\frac{2 \mathrm{t}}{3}\right)+\mathrm{C}=\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{3}\right)+\mathrm{C}$

Hence, the correct answer is option: 'C'.

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