Find the following integral:int frac{1}{1+tan | vedclass

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

Question

(A) Let $I = \int \frac{1}{1+	an x} d x$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int \frac{\cos x}{\cos x + \sin x} d x$.Multiply the

Vedclass · Accepted Answer

$\frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C$

Vedclass · Answer

(A) Let $I = \int \frac{1}{1+	an x} d x$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int \frac{\cos x}{\cos x + \sin x} d x$.Multiply the numerator and denominator by $2$ and adjust the numerator:$I = \frac{1}{2} \int \frac{2\cos x}{\cos x + \sin x} d x = \frac{1}{2} \int \frac{(\cos x + \sin x) + (\cos x - \sin x)}{\cos x + \sin x} d x$.Split the integral:$I = \frac{1}{2} \int 1 d x + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} d x$.For the second integral,let $u = \cos x + \sin x$,then $du = (-\sin x + \cos x) d x$.Thus,$I = \frac{x}{2} + \frac{1}{2} \int \frac{1}{u} du = \frac{x}{2} + \frac{1}{2} \log |u| + C$.Substituting back $u = \cos x + \sin x$,we get:$I = \frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C$.

Find the following integral:
$\int \frac{1}{1+\tan x} d x$

Similar Questions

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

If $\int {({x^3} - 2{x^2} + 5){e^{3x}}\,dx} = e^{3x} (Ax^3 + Bx^2 + Cx + D) + K$,then the statement which is incorrect is

$\int \frac{dx}{\tan x + \cot x + \sec x + \csc x} =$

If $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$,then $f\left(\frac{2 \pi}{3}\right) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the following integral:$\int \frac{1}{1+\tan x} d x$