definite integral hirap po talaga hindi ko masagot

Submitted by Jeanvill Palad… on

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Integration

Jhun Vert Tue, 02/09/2016 - 23:08

$\displaystyle \int_0^{1/3} \dfrac{dx}{(4 - 9x^2)^2}$

      $\displaystyle = \int_0^{1/3} \dfrac{dx}{[ \, 2^2 - (3x)^2 \, ]^2}$

Let
3x = 2 sin θ
3 dx = 2 cos θ dθ
dx = (2/3) cos θ dθ

When
x = 0, θ = 0,
x = 1/3, θ = 30° = π/6
 

Hence,
$\displaystyle \int_0^{1/3} \dfrac{dx}{(4 - 9x^2)^2}$

      $\displaystyle = \int_0^{\pi/6} \dfrac{\frac{2}{3} \cos \theta \, d\theta}{[ \, 2^2 - (2\sin \theta)^2 \, ]^2}$

      $\displaystyle = \dfrac{1}{24} \int_0^{\pi/6} \dfrac{\cos \theta \, d\theta}{(1 - \sin^2 \theta)^2}$

      $\displaystyle = \dfrac{1}{24} \int_0^{\pi/6} \dfrac{\cos \theta \, d\theta}{\cos^4 \theta}$

      $\displaystyle = \dfrac{1}{24} \int_0^{\pi/6} \dfrac{d\theta}{\cos^3 \theta}$

      $\displaystyle = \dfrac{1}{24} \int_0^{\pi/6} \sec^3 \theta \, d\theta$

For

$\displaystyle \int \sec^3 \theta \, d\theta$

Let
u = sec θ
du = sec θ tan θ dθ

dv = sec2 θ dθ
v = tan θ
 

$\displaystyle \int \sec^3 \theta \, d\theta$

      $\displaystyle = \sec \theta \, \tan \theta - \int \tan \theta (\sec \theta \, \tan \theta) \, d\theta$

      $\displaystyle = \sec \theta \, \tan \theta - \int \tan^2 \theta \sec \theta \, d\theta$

      $\displaystyle = \sec \theta \, \tan \theta - \int (\sec^2 \theta - 1) \sec \theta \, d\theta$

      $\displaystyle = \sec \theta \, \tan \theta - \int \sec^3 \theta \, d\theta + \int \sec \theta \, d\theta$

$\displaystyle 2\int \sec^3 \theta \, d\theta = \sec \theta \, \tan \theta + \int \sec \theta \, d\theta$

$\displaystyle 2\int \sec^3 \theta \, d\theta = \sec \theta \, \tan \theta + \ln (\sec \theta + \tan \theta)$

$\displaystyle \int \sec^3 \theta \, d\theta = \frac{1}{2}\sec \theta \, \tan \theta + \frac{1}{2}\ln (\sec \theta + \tan \theta)$

 

Hence,
$\displaystyle \int_0^{1/3} \dfrac{dx}{(4 - 9x^2)^2}$

      $\displaystyle = \dfrac{1}{24} \int_0^{\pi/6} \sec^3 \theta \, d\theta$

      $= \dfrac{1}{24} \left[ \dfrac{1}{2}\sec \theta \, \tan \theta + \dfrac{1}{2}\ln (\sec \theta + \tan \theta) \right]_0^{\pi/6}$

      $= \dfrac{1}{48} \left[ \sec \theta \, \tan \theta + \ln (\sec \theta + \tan \theta) \right]_0^{\pi/6}$

      $= \dfrac{1}{48} \left[ \sec \dfrac{\pi}{6} \, \tan \dfrac{\pi}{6} + \ln \left( \sec \dfrac{\pi}{6} + \tan \dfrac{\pi}{6} \right) \right] \\ ~ ~ ~ ~ ~ - \dfrac{1}{48} \left[ \sec 0 \, \tan 0 + \ln \left (\sec 0 + \tan 0 \right) \right]$

      $= \dfrac{1}{48} \left[ \dfrac{2}{\sqrt{3}} \left( \dfrac{\sqrt{3}}{3} \right) + \ln \left( \dfrac{2}{\sqrt{3}} + \dfrac{\sqrt{3}}{3} \right) \right] - \dfrac{1}{48} \left[ 1(0) + \ln \left( 1 + 0 \right) \right]$

      $= \dfrac{1}{48} \left[ \dfrac{2}{3} + \ln \left( \sqrt{3} \right) \right]$

      $= 0.0253$