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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) माना $I = \int \frac{dx}{	an x + \cot x + \sec x + \csc x}$.$= \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$अंश और हर को $(\sin x + \cos x - 1)$ से गुणा करने पर:$= \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

$\int \sqrt{\frac{2+x}{2-x}} \, dx$ का मान क्या है?

यदि $\int e^x \cos x \, dx = \frac{e^x}{2}(\cos x + \sin x)$ और $\int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx = \frac{f(x)}{24}[\cos (g(x)) + \sin (g(x))] + c$ है, तो $g(1) =$

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ का मान ज्ञात कीजिए।

$\int \frac{1}{3 \cos x - 4 \sin x + 5} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module