3+4 ^2x dx= - Vidyadip

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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