sec-2-x-3sin-2-x-1-dx-

Question Number 156305 by cortano last updated on 10/Oct/21

\int \frac{\sec^{2} (x)}{3 \sin^{2} (x) - 1} dx

Answered by puissant last updated on 10/Oct/21

Ω = \int \frac{{s e c}^{2} x}{3 {s i n}^{2} x - 1} d x = \int \frac{{s e c}^{4} x}{3 {t a n}^{2} x - {s e c}^{2} x} d x

u = t a n x \to d u = {s e c}^{2} x d x = (1 + u^{2}) d u

\Rightarrow Ω = \int \frac{1 + u^{2}}{3 u^{2} - (1 + u^{2})} d u = \int \frac{1 + u^{2}}{2 u^{2} - 1} d u

Ω = \int \frac{1}{2 u^{2} - 1} d u + \int \frac{u^{2}}{2 u^{2} - 1} d u

= \frac{1}{2} \int \frac{d u}{u^{2} - \frac{1}{2}} + \frac{1}{2} \int \frac{2 u^{2} - 1 + 1}{2 u^{2} - 1} d u

= \frac{1}{2} \int \frac{d u}{u^{2} - \frac{1}{2}} + \frac{1}{4} \int \frac{d u}{u^{2} - \frac{1}{2}} + \frac{1}{2} \int d u = \frac{3}{4} \int \frac{d u}{u^{2} - \frac{1}{2}} + \frac{1}{2} \int d u

= - \frac{6}{4} \int \frac{d u}{1 - (\frac{u}{\sqrt{2}})} - \frac{u}{2} = - \frac{6 \sqrt{2}}{4} a r g t a n h (\sqrt{2} u) - \frac{u}{2} + C . .

∴∵ Ω = - \frac{3 \sqrt{2}}{2} a r g t a n h (\sqrt{2} t a n x) - \frac{t a n x}{2} + C

Commented by cortano last updated on 10/Oct/21

ok . i got in \tan x form .

Commented by puissant last updated on 10/Oct/21

O k e y \dots

Commented by puissant last updated on 10/Oct/21

M r C o r t a n o, a n o t h e r i m p o r t a n t

f o r m u l a :

a r g t h (x) + C = \frac{1}{2} l n (\frac{x + 1}{x - 1}) + C = \int \frac{1}{x^{2} - 1} d x . .

Commented by cortano last updated on 10/Oct/21

thank you sir

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