Mathematics

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


ANSWER

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


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
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