int frac{dx}{tan x + cot x + sec x + csc x} = | vedclass

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

Question

(B) ધારો કે $I = \int \frac{dx}{	an x + \cot x + \sec x + \csc x}$.$= \int \frac{dx}{\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\

Vedclass · Accepted Answer

$\frac{1}{2} (\sin x - \cos x - x) + c$

Vedclass · Answer

(B) ધારો કે $I = \int \frac{dx}{	an x + \cot x + \sec x + \csc x}$.$= \int \frac{dx}{\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x}ight)}$$= \int \frac{\sin x \cos x}{\sin^{2} x + \cos^{2} x + \sin x + \cos x} dx$$= \int \frac{\sin x \cos x}{1 + \sin x + \cos x} dx$અંશ અને છેદને $(\sin x + \cos x - 1)$ વડે ગુણતા:$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{(\sin x + \cos x)^{2} - 1} dx$$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1} dx$$= \int \frac{\sin x \cos x (\sin x + \cos x - 1)}{2 \sin x \cos x} dx$$= \frac{1}{2} \int (\sin x + \cos x - 1) dx$$= \frac{1}{2} (-\cos x + \sin x - x) + c$.

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

Similar Questions

$\int \frac{d x}{4 \sin x+3 \cos x}=$

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

$\int \frac{x^2-2}{x^3 \sqrt{x^2-1}} d x=$

$\int \sqrt{3-2x-x^2} \, dx = $ . . . . . . $+ C$.

$\int \frac{x^4+1}{1+x^6} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module