Mathematics

# Evaluate the following integrals$\int { \cfrac { 1 }{ \sin { x } +\cos { x } } } dx$

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

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

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

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

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

So $I=-\displaystyle\int \dfrac{2 d t}{ t^2-2 t-1}$

$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)-1}$

$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)+(1)^2-1-(1)^2}$

$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-2}$                    $(\because (a-b)^2=a^2-2 a b+b^2\ )$

$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-(\sqrt{2})^2}$

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

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

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

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