Question
Fill in the blanks:
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}=$ ________.
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}=$ ________.
✓
Answer
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}=-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
Solution:
Let $\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}$
Put $\cos\text{x}=\text{t}\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{3+4\text{t}^2}=-\frac{1}{4}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{2}\Big)^2+\text{t}^2}$
$=-\frac{1}{4}\cdot\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\text{t}}{\sqrt{3}}+\text{C}=-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
Solution:
Let $\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}$
Put $\cos\text{x}=\text{t}\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{3+4\text{t}^2}=-\frac{1}{4}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{2}\Big)^2+\text{t}^2}$
$=-\frac{1}{4}\cdot\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\text{t}}{\sqrt{3}}+\text{C}=-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
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