Mathematics

$\displaystyle \int\frac{d{x}}{\cos x-\sin x}=$

$\displaystyle \frac{1}{\sqrt{2}}\log|\tan(\frac{x}{2}-\frac{3\pi}{8})|+c$

SOLUTION
$\displaystyle \int \frac{dx}{(cos\ x-sin\ x)}= \int \frac{\left ( \begin{matrix}1+tan^{2}\ ^{x}/_{2}\\dx \end{matrix} \right )}{\left ( 1-tan^{2}\ ^{x}/_{2} \right )-(2tan\ ^{x}/_{2})}$
$\displaystyle \int \frac{sec^{2}\ ^{x}/_{2}}{\left [ 1-tan^{2}\ ^{x}/_{2}-2tan\ ^{x}/_{2} \right ]}dx$
Let, $\ tan^{x}/_{2}= t$
$\displaystyle ^{1}/_{2}\ sec^{2}\ ^{x}/_{2}\cdot dx= 2dt$
$\displaystyle \int \frac{2dt}{(1-t^{2}-2t)}= -2\int \frac{dt}{(t^{2}+2t-1)}$
$\displaystyle -2\int \frac{dt}{(t+1)^{2}-(\sqrt{2})^{2}}$
$\displaystyle \frac{-2}{2\sqrt{2}}\ log\left | \frac{(t+1)-\sqrt{2}}{(t+1)+\sqrt{2}} \right |$
$\displaystyle = \frac{2}{2\sqrt{2}}\ log\left | \frac{t-(1+\sqrt{2})}{t+(1-\sqrt{2})} \right |+c$
$\displaystyle =\ ^{1}/_{\sqrt{2}}\ log\left ( tan\left ( ^{x}/_{2}-^{3\pi }/_{8} \right ) \right )+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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