dx-tgx-tgx-

Question Number 51122 by behi83417@gmail.com last updated on 24/Dec/18

\int \frac{dx}{tgx - \sqrt{tgx}} = ?

Commented by MJS last updated on 24/Dec/18

long way \dots

start with

t = \sqrt{\tan x} \to d x = \frac{2 t}{t^{4} + 1} d t

\Rightarrow 2 \int \frac{d t}{(t + 1) (t^{4} + 1)} and continue with decomposing \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

\int \frac{d x}{t a n x - \sqrt{t a n x}}

t^{2} = t a n x 2 t d t = {s e c}^{2} x d x

\frac{2 t}{1 + t^{4}} d t = d x

\int \frac{d x}{\sqrt{t a n x} (\sqrt{t a n x} - 1)}

\int \frac{\sqrt{t a n x} - (\sqrt{t a n x} - 1)}{\sqrt{t a n x} (\sqrt{t a n x} - 1)} d x

\int \frac{d x}{\sqrt{t a n x} - 1} - \int \frac{d x}{\sqrt{t a n x}}

I_{1} - I_{2}

I_{2} = \int \frac{2 t}{1 + t^{4}} \times \frac{1}{t} d t

= \int \frac{2 d t}{1 + t^{4}}

\int \frac{\frac{2}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t

\int \frac{(1 + \frac{1}{t^{2}}) - (1 - \frac{1}{t^{2}})}{(t^{2} + \frac{1}{t^{2}})} d t

\int \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2} - \int \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - 2}

\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}) + c_{1}

\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{\sqrt{t a n x} - \frac{1}{\sqrt{t a n x}}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{\sqrt{t a n x} + \frac{1}{\sqrt{t a n x}} - \sqrt{2}}{\sqrt{t a n x} + \frac{1}{\sqrt{t a n x}} + \sqrt{2}}) + c_{1}

I_{1} = \int \frac{d x}{\sqrt{t a n x} - 1}

= \int \frac{2 t d t}{1 + t^{4}} \times \frac{1}{t - 1}

2 \int \frac{t - 1 + 1}{(t^{4} + 1) (t - 1)} d t

2 \int \frac{d t}{t^{4} + 1} + 2 \int \frac{d t}{(t - 1) (t^{4} + 1)} \leftarrow S O L V E I T \dots

I_{2} + 2 I_{3}

Commented by behi83417@gmail.com last updated on 25/Dec/18

t h a n k y o u s i r t a n m a y .