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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals1 Mark
MCQ
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}=$
  • A
    $\log(3+4\cos^2\text{x})+\text{C}$
  • B
    $\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
  • ✓
    $-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
  • D
    $\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
✓

Answer

Correct option: C.
$-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}$
Put $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\text{I}=\int\frac{-\text{dt}}{3+4\text{t}^2}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

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