Question

$\int \frac{dx}{\cos x-\sin x}=$

• A. $\frac{1}{\sqrt{2}}\log |\tan (\frac{x}{2}-\frac{\pi }{8})|+c$
• B. $\frac{1}{\sqrt{2}}\log |\tan (\frac{x}{2}+\frac{3\pi }{8})|+c$
• C. $\frac{1}{\sqrt{2}}\log |\tan (\frac{x}{2}-\frac{3\pi }{8})|+c$
• D. $\frac{1}{\sqrt{2}}\log |\cot \left(\frac{x}{2}\right)|+c$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}\log |\tan (\frac{x}{2}-\frac{3\pi }{8})|+c$

$\int \frac{dx}{(cosx-sinx)}=\int \frac{\left(\begin{array}{c}1+ta{n}^2{}^x{/}_2\\ dx\end{array}\right)}{(1-ta{n}^2{}^x{/}_2)-\left(2tan{}^x{/}_2\right)}$
$\int \frac{se{c}^2{}^x{/}_2}{[1-ta{n}^2{}^x{/}_2-2tan{}^x{/}_2]}dx$
Let, $ta{n}^x{/}_2=t$
${}^1{/}_{2}se{c}^2{}^x{/}_2\cdot dx=2dt$
$\int \frac{2dt}{(1-t^2-2t)}=-2\int \frac{dt}{(t^2+2t-1)}$
$-2\int \frac{dt}{(t+1{)}^2-(\sqrt{2}{)}^2}$
$\frac{-2}{2\sqrt{2}}log\left|\frac{(t+1)-\sqrt{2}}{(t+1)+\sqrt{2}}\right|$
$=\frac{2}{2\sqrt{2}}log\left|\frac{t-(1+\sqrt{2})}{t+(1-\sqrt{2})}\right|+c$
$={}^1{/}_{\sqrt{2}}log\left(tan\left({}^x{/}_2{-}^{3\pi }{/}_8\right)\right)+c$