int frac{dx}{sin x + cos x} = | vedclass

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

Question

(C) To evaluate the integral $I = \int \frac{dx}{\sin x + \cos x}$,we multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{dx}{\sin x + \cos x}$,we multiply and divide by $\sqrt{2}$:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x}$Using the identity $\sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x$,we get:$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\sin(x + \frac{\pi}{4})}$$I = \frac{1}{\sqrt{2}} \int \csc(x + \frac{\pi}{4}) dx$Using the standard integral formula $\int \csc 	heta d	heta = \log |	an(\frac{	heta}{2})| + c$,we have:$I = \frac{1}{\sqrt{2}} \log \left| 	an \left( \frac{x + \frac{\pi}{4}}{2} ight) ight| + c$$I = \frac{1}{\sqrt{2}} \log 	an \left( \frac{x}{2} + \frac{\pi}{8} ight) + c$

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

Similar Questions

$\int \sqrt{5-2x+x^2} dx$ is equal to

If $f^{\prime}(x)=a \sin x+b \cos x$,$f^{\prime}(0)=4$,$f(0)=3$ and $f\left(\frac{\pi}{2}\right)=5$,then $f(x)=$

$\int \sin^{-1}(\cos x) \, dx = $

$ \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx $ is equal to

$\int (\tan^7 x + \tan x) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module