find ∫ (dx/(3+tan^2 x))


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Question Number 42430 by maxmathsup by imad last updated on 25/Aug/18

$f i n d \int \frac{d x}{3 + {t a n}^{2} x}$
Commented by maxmathsup by imad last updated on 25/Aug/18

$l e t A = \int \frac{d x}{3 + {t a n}^{2} x} c h a n g e m e n t t a n x = \sqrt{3} t g i v e x = a r c t a n (\sqrt{3} t)$ $A = \int \frac{1}{3 (1 + t^{2})} \frac{\sqrt{3}}{1 + 3 t^{2}} d t = \frac{1}{\sqrt{3}} \int \frac{d t}{(t^{2} + 1) (3 t^{2} + 1)} l e t f i n d a a n d b /$ $F (t) = \frac{1}{(t^{2} + 1) (3 t^{2} + 1)} = \frac{a}{t^{2} + 1} + \frac{b}{3 t^{2} + 1}$ ${l i m}_{t \to + \infty} t^{2} F (t) = 0 = a + \frac{b}{3} \Rightarrow 3 a + b = 0 \Rightarrow b = - 3 a \Rightarrow$ $F (t) = \frac{a}{t^{2} + 1} - \frac{3 a}{3 t^{2} + 1}$ $F (0) = 1 = - 2 a \Rightarrow a = - \frac{1}{2} \Rightarrow b = \frac{3}{2} \Rightarrow F (t) = - \frac{1}{2 (t^{2} + 1)} + \frac{3}{2 (3 t^{2} + 1)} \Rightarrow$ $A = \int F (t) d t = - \frac{1}{2} \int \frac{d t}{t^{2} + 1} + \frac{3}{2} \int \frac{d t}{1 + 3 t^{2}} b u t \int \frac{d t}{t^{2} + 1} = a r c t a n t$ $\int \frac{d t}{1 + 3 t^{2}} =_{\sqrt{3} t = u} \frac{1}{\sqrt{3}} \int \frac{d u}{1 + u^{2}} = \frac{1}{\sqrt{3}} a r c t a n (\sqrt{3} t) \Rightarrow$ $A = - \frac{1}{2} a r c t a n t + \frac{\sqrt{3}}{2} a r c t a n (\sqrt{3} t) + c$ $= - \frac{1}{2} a r c t a n (\frac{t a n x}{\sqrt{3}}) + \frac{\sqrt{3}}{2} x + c .$
Commented by maxmathsup by imad last updated on 25/Aug/18

$A = \frac{1}{\sqrt{3}} \int F (t) d t \Rightarrow A = - \frac{1}{2 \sqrt{3}} a r c t a n (\frac{t a n x}{\sqrt{3}}) + \frac{x}{2} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

	$\int \frac{d x}{3 + \frac{1 - c o s 2 x}{1 + c o s 2 x}}$ $\int \frac{1 + c o s 2 x}{3 + 3 c o s 2 x + 1 - c o s 2 x} d x$ $\int \frac{1 + c o s 2 x}{4 + 2 c o s 2 x} d x$ $\frac{1}{2} \int \frac{2 + c o s 2 x - 1}{2 + c o s 2 x} d x$ $\frac{1}{2} \int d x - \frac{1}{2} \int \frac{d x}{2 + c o s 2 x}$ $\frac{\frac{1}{2} \int d x - \frac{1}{2} \int \frac{1 + {t a n}^{2} x}{2 + 2 {t a n}^{2} x + 1 - {t a n}^{2} x} d x}{}$ $\frac{1}{2} \int d x - \frac{1}{2} \int \frac{{s e c}^{2} x}{3 + {t a n}^{2} x} d x$ $\frac{1}{2} \int d x - \frac{1}{2} \int \frac{d (t a n x)}{3 + {t a n}^{2} x} d x \to \int \frac{d x}{x^{2} + a^{2}} f o r m u l a$ $\frac{x}{2} - \frac{1}{2} \times \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{t a n x}{\sqrt{3}}) + c$
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