int frac{dx}{cos x - sin x} is equal to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{dx}{\cos x - \sin x}$.Multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \cos x - \fr

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{3\pi}{8} ight) ight| + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\cos x - \sin x}$.Multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}$Using $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos(x + \frac{\pi}{4})}$$I = \frac{1}{\sqrt{2}} \int \sec(x + \frac{\pi}{4}) dx$Using the formula $\int \sec 	heta d	heta = \log |	an(\frac{	heta}{2} + \frac{\pi}{4})| + c$:$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x + \frac{\pi}{4}}{2} + \frac{\pi}{4} ight) ight| + c$$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{\pi}{8} + \frac{2\pi}{8} ight) ight| + c$$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x}{2} + \frac{3\pi}{8} ight) ight| + c$.

$\int \frac{dx}{\cos x - \sin x}$ is equal to

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = e^x + \cos x + x + \tan x$ is

$\int {\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2} dx = $

$\int \sec x \, dx = $

$\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx = $

If $\int \frac{x+5}{x^2+4x+5} dx = a \log(x^2+4x+5) + b \tan^{-1}(x+k) + C$, then $(a, b, k)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module