First slide

Methods of integration

Question

$Evaluate \int \sqrt{\tan θ} d θ$ = $\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) + \frac{1}{2 \sqrt{2}} \log |\frac{A}{B}| + C$

Difficult

A

$|\frac{\tan θ - \sqrt{2 \tan θ} + 1}{\tan θ + \sqrt{2 \tan θ} + 1}|$
B

$|\frac{\tan θ - 2 \sqrt{\tan θ} + 1}{\tan θ + 2 \sqrt{\tan θ} + 1}|$
C

$|\frac{\tan θ + \sqrt{2 \tan θ} + 1}{\tan θ - \sqrt{2 \tan θ} + 1}|$
D

None of these

Solution

$Let I = \int \sqrt{\tan θ} d θ Let \tan θ = x^{2} . Then, d (\tan θ) = d (x^{2}) or \sec^{2} θ d θ = 2 x d x or \begin{array}{l} d θ = \frac{2 x \cdot d x}{\sec^{2} θ} = \frac{2 x d x}{1 + \tan^{2} θ} = \frac{2 x d x}{1 + x^{4}} \\ I = \int \sqrt{x^{2}} \times \frac{2 x d x}{1 + x^{4}} = \int \frac{2 x^{2}}{x^{4} + 1} d x \\ = \int \frac{2}{x^{2} + 1 / x^{2}} d x \\ = \int \frac{1 + 1 / x^{2} + 1 - 1 / x^{2}}{x^{2} + 1 / x^{2}} d x \\ = \int \frac{1 + 1 / x^{2}}{x^{2} + 1 / x^{2}} d x + \int \frac{1 - 1 / x^{2}}{x^{2} + 1 / x^{2}} d x \\ = \int \frac{1 + 1 / x^{2}}{(x - 1 / x)^{2} + 2} d x + \int \frac{1 - 1 / x^{2}}{(x + 1 / x)^{2} - 2} d x \\ p u t t i n g x - \frac{1}{x} = u i n t h e f i r s t i n t e g r a l a n d x + \frac{1}{x} = v in the second integral we get \\ \begin{array}{l} I = \int \frac{d u}{u^{2} + (\sqrt{2})^{2}} + \int \frac{d v}{v^{2} - (\sqrt{2})^{2}} \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + \frac{1}{2 \sqrt{2}} \log |\frac{v - \sqrt{2}}{v + \sqrt{2}}| + C \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x - 1 / x}{\sqrt{2}}) + \frac{1}{2 \sqrt{2}} \log |\frac{x + 1 / x - \sqrt{2}}{x + 1 / x + \sqrt{2}}| + C \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{x \sqrt{2}}) + \frac{1}{2 \sqrt{2}} \log |\frac{x^{2} - x \sqrt{2} + 1}{x^{2} + x \sqrt{2} + 1}| + C \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan θ - 1}{\sqrt{2 \tan θ}}) + \frac{1}{2 \sqrt{2}} \log |\frac{\tan θ - \sqrt{2 \tan θ} + 1}{\tan θ + \sqrt{2 \tan θ} + 1}| + C \end{array} \end{array}$

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