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

यदि $\int \frac{x-\sin x}{1+\cos x} dx = x \tan \left(\frac{x}{2}\right) + p \log \left|\sec \left(\frac{x}{2}\right)\right| + C$ है,तो $p$ का मान ज्ञात कीजिए।

यदि $\int \frac{d x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(f(x))}{\sqrt{5}}+c$,(जहाँ $c$ समाकलन का स्थिरांक है),तो $f(\pi / 4)$ का मान ज्ञात कीजिए:

यदि $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C, (a>0, b>0)$, तो $\int \frac{d x}{b+a \cos x}=$

यदि $\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{2\sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) - \frac{1}{2} \log g(x) + C$ है,तो $g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module