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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
If $\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C},$ then:
  • ✓
    $\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
  • B
    $\text{A}=\frac{1}{3},\text{B}=\frac{2}{3}$
  • C
    $\text{A}=-\frac{2}{3},\text{B}=\frac{5}{3}$
  • D
    $\text{A}=\frac{1}{3},\text{B}=-\frac{5}{3}$
✓

Answer

Correct option: A.
$\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
$\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C}\ ...(\text{i})$
Considering the $\text{LHS}$ of $eq. (1)$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\int\frac{1}{5+\frac{8\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{dx}$
$\Rightarrow\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{dx}$
$\Rightarrow\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{dx}\dots(2)$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\sec^2\frac{\text{x}}{2}\times\frac{1}{2}\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{2dt}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+\big(\frac{4}{5}\big)^2-\big(\frac{4}{5}\big)^2+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+1-\frac{16}{25}}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+\big(\frac{3}{5}\big)^2}$
$\Rightarrow\frac{2}{5}\times\frac{2}{3}\tan^{-1}\bigg(\frac{\text{t}+\frac{4}{5}}{\frac{3}{5}}\bigg)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}+4}{3}\Big)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5}{3}\tan\frac{\text{x}}{2}+\frac{4}{3}\Big)+\text{C}\ ...\text{(ii)}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$
Comparing $eq. (ii)$ with the $\text{RHS}$ of $eq. (i)$ we get,
$\therefore\ \text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$

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