Integrate the functions in Exercises: - Vidyadip

Question

Integrate the functions in Exercises:
$\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}$

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Answer

$\text{Let I}=\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{ dx}=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}.\cos\text{x}}\text{ dx} $ $=\int\frac{\sqrt{\tan\text{x}}}{{\tan\text{x}\cos^2\text{x}}}\text{ dx} =\int\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)}$ Putting $\tan\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}\text{ dx = dt}$ $\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{dt}=\frac{\text{t}^{^1/_2}}{^1/_2}+\text{c}$$=2\sqrt{\text{t}}+\text{c}=2\sqrt{\tan\text{x}}+\text{c}$

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