Mathematics

# Evaluate the following integrals:$\int { \cfrac { 1 }{ 4\cos { x } -1 } } dx\quad$

##### SOLUTION
Let $\text{I}=\displaystyle\int \dfrac{1}{4\cos x-1} d x$

$=\displaystyle\int \dfrac{1}{4\bigg(\dfrac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}\bigg)-1}d x$                       $\bigg(\because \cos 2 A=\dfrac{1-\tan^2A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{-\bigg(1+\tan^2 \dfrac{x}{2}\bigg)+4\bigg(1-\tan^2 \dfrac{x}{2}\bigg)}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{3-5\tan^2 \dfrac{x}{2}}d x$               $(\because \text{sec}^2 A-\tan^2 A=1)$

Let $t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

So $I=\displaystyle\int \dfrac{2 d t}{3-5 t^2}$

$=-\dfrac{2}{5}\displaystyle\int \dfrac{d t}{t^2-\bigg(\sqrt{\dfrac{3}{5}}\bigg)^2}$

$=-\dfrac{2}{5}\times \dfrac{1}{2\sqrt{\dfrac{3}{5}}}\log \bigg| \dfrac{t-\sqrt{\dfrac{3}{5}}}{t+\sqrt{\dfrac{3}{5}}}\bigg|+c$                                        $\bigg(\because \displaystyle\int \dfrac{d x}{x^2-a^2}=\dfrac{1}{2 a}\log \bigg|\dfrac{x-a}{x+a}\bigg|+c\ \bigg)$

$=-\dfrac{\sqrt{5}}{5\sqrt{3}}\log \bigg| \dfrac{\sqrt{5} t-\sqrt{3}}{\sqrt{5} t+\sqrt{3}}\bigg|+c$

$=-\dfrac{1}{\sqrt{15}}\log \bigg|\dfrac{\sqrt{5}\tan \dfrac{x}{2}-\sqrt{3}}{\sqrt{5}\tan \dfrac{x}{2}+\sqrt{3}}\bigg|+c$                                       $\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.