Question

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

A
12log|tan(x2π8)|+c
B
12log|tan(x2+3π8)|+c
C
12log|tan(x23π8)|+c
D
12log|cot(x2)|+c

Solution

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

Suggest Corrections

0

Similar questions
View More