Question-117511

Question Number 117511 by Canovas last updated on 12/Oct/20

Answered by bemath last updated on 12/Oct/20

\int \frac{dx}{\cos^{2} x (4 - 9 \tan^{2} x)} = \int \frac{\sec^{2} x}{4 - 9 \tan^{2} x} dx

= \int \frac{d (\tan x)}{4 - 9 \tan^{2} x} = \int \frac{d φ}{4 - 9 φ^{2}}

= \int \frac{d φ}{(2 + 3 φ) (2 - 3 φ)} = \frac{1}{4} \int \frac{1}{2 - 3 φ} + \frac{1}{2 + 3 φ} d φ

= \frac{1}{4} {- \frac{1}{3} \ln (2 - 3 φ) + \frac{1}{3} \ln (2 + 3 φ)} + c

= \frac{1}{12} \ln ∣ \frac{2 + 3 φ}{2 - 3 φ} ∣ + c

= \frac{1}{12} \ln ∣ \frac{2 + 3 \tan x}{2 - 3 \tan x} ∣ + c

Commented by Canovas last updated on 12/Oct/20

R e a l l y I {h a v e n}^{'} t u n d e r s t o o d s i r

Commented by bobhans last updated on 12/Oct/20

it is by substitution sir .

φ = \tan x so d φ = \sec^{2} x dx

Answered by 1549442205PVT last updated on 12/Oct/20

F = \int \frac{dx}{{4 \cos}^{2} x - {9 \sin}^{2} x} = \int \frac{\frac{1}{\cos^{2} x}}{4 - {9 \tan}^{2} x} dx

= - \int \frac{1 + \tan^{2} x}{{9 \tan}^{2} x - 4} dx . Put tanx = t

\Rightarrow dt = (1 + t^{2}) dx

F = - \int \frac{(1 + t^{2}) dt}{(1 + t^{2}) ({9 t}^{2} - 4)} = + \int \frac{dt}{{9 t}^{2} - 4}

= - \frac{1}{9} \int \frac{dt}{t^{2} - \frac{4}{9}} = - \frac{1}{9} \times \frac{3}{4} \int (\frac{1}{t - \frac{2}{3}} - \frac{1}{t + \frac{2}{3}}) dt

= - \frac{1}{12} \ln ∣ \frac{3 t - 2}{3 t + 2} ∣ + C = \frac{1}{12} \ln ∣ \frac{3 tanx + 2}{3 tanx - 2} ∣ + C

Answered by mathmax by abdo last updated on 12/Oct/20

I = \int \frac{dx}{{4 \cos}^{2} x - {9 \sin}^{2} x} \Rightarrow I = \int \frac{dx}{4 \frac{1 + \cos (2 x)}{2} - 9 \frac{1 - \cos (2 x)}{2}}

= \int \frac{dx}{2 + 2 \cos (2 x) - \frac{9}{2} + \frac{9}{2} \cos (2 x)} = \int \frac{2 dx}{4 + 4 \cos (2 x) - 9 + 9 \cos (2 x)}

= \int \frac{2 dx}{13 \cos (2 x) - 5} =_{2 x = t} \int \frac{dt}{13 cost - 5} =_{\tan (\frac{t}{2}) = u}

\int \frac{2 du}{(1 + u^{2}) (13 \frac{1 - u^{2}}{1 + u^{2}} - 5)} = \int \frac{2 du}{13 (1 - u^{2}) - 5 (1 + u^{2})}

= \int \frac{2 du}{13 - {13 u}^{2} - 5 - {5 u}^{2}} = \int \frac{2 du}{8 - {18 u}^{2}} = \int \frac{du}{4 - {9 u}^{2}}

= \int \frac{du}{(2 - 3 u) (2 + 3 u)} = \frac{1}{4} \int (\frac{1}{2 - 3 u} + \frac{1}{2 + 3 u}) du = \frac{1}{4} \ln ∣ 4 - {9 u}^{2} ∣ + c

= \frac{1}{4} \ln ∣ 4 - {9 \tan}^{2} (x) ∣ + c

Commented by bobhans last updated on 13/Oct/20

how do you get \frac{1}{4} \int (\frac{1}{2 - 3 u} + \frac{1}{2 + 3 u}) du

= \frac{1}{4} \ln ∣ 4 - {9 u}^{2} ∣ ? it not correct

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