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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
MCQ
$\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}=$
  • ✓
    $\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
  • B
    $\frac{\pi}{\text{ab}}$
  • C
    $\frac{\pi}{\text{a}^2-\text{b}^2}$
  • D
    $({\text{a}+\text{b}})\pi$
✓

Answer

Correct option: A.
$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
We have,
$\text{I}=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{\text{a}\Big(1+\tan^2\frac{\text{x}}{2}\Big)+\Big(1-\tan^2\frac{\text{x}}{2}\Big)}\text{ dx}$
$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$
$=\int\limits^\pi_0\frac{\sec^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$
putting $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow \frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\Rightarrow\sec^2\frac{\text{x}}{2}\text{dx}=2\text{ dt}$
when $\text{x}\rightarrow0;\text{ t}\rightarrow0$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{2\text{dt}}{(\text{a}+\text{b})+(\text{a}-\text{b})\text{t}^2}$
$=\frac{2}{\text{a}-\text{b}}\int\limits^\pi_0\frac{1}{\big(\frac{\text{a}+\text{b}}{\text{a}-\text{b}}\big)+\text{t}^2}\text{dt}$
$=\frac{2}{(\text{a}-\text{b})}\int\limits^\infty_0\frac{1}{\Big(\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}\Big)^2+\text{t}^2}\text{ dt}$
$=\frac{2}{(\text{a}-\text{b})}\times\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$ $\Bigg[\tan^{-1}\frac{\text{t}}{\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}}\Bigg]^\infty_0$
$=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\Big[\frac{\pi}{2}\Big]$
$=\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

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