int frac{dx}{7 + 5cos x} = | vedclass

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

Question

(A) माना $I = \int \frac{dx}{7 + 5\cos x}$.$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ सूत्र का उपयोग करने पर:$I = \int \frac{dx}{7 + 5 \left(

Vedclass · Accepted Answer

$\frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{1}{\sqrt{6}} 	an \frac{x}{2} ight) + c$

Vedclass · Answer

(A) माना $I = \int \frac{dx}{7 + 5\cos x}$.$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ सूत्र का उपयोग करने पर:$I = \int \frac{dx}{7 + 5 \left( \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight)} = \int \frac{(1 + 	an^2(x/2)) dx}{7(1 + 	an^2(x/2)) + 5(1 - 	an^2(x/2))}$$I = \int \frac{\sec^2(x/2) dx}{7 + 7	an^2(x/2) + 5 - 5	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{12 + 2	an^2(x/2)}$$I = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{6 + 	an^2(x/2)}$माना $	an(x/2) = t$,तब $\frac{1}{2} \sec^2(x/2) dx = dt$.$I = \int \frac{dt}{t^2 + (\sqrt{6})^2} = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{t}{\sqrt{6}} ight) + c$$t = 	an(x/2)$ का मान रखने पर:$I = \frac{1}{\sqrt{6}} 	an^{-1} \left( \frac{	an(x/2)}{\sqrt{6}} ight) + c$.

$\int \frac{dx}{7 + 5\cos x} = $

Similar Questions

समाकलन $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$ का मान ज्ञात कीजिए।

$\int \frac{\log (\cot x)}{\sin 2 x} \,d x=$

$\int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx = $

यदि $\int \frac{\sin 2x \, dx}{\sin^4 x + \cos^4 x} = \tan^{-1}(f(x)) + c$ है,तो $f\left(\frac{\pi}{3}\right) = $

यदि $\int \frac{dx}{5 + 4\cos x} = \lambda \tan^{-1} \left( m \tan \frac{x}{2} \right) + C$ है,तो:

Vedclass Test Series

Exam Paper Generator

Online Exam Module