यदि int left( frac{1-5 cos^{2}x}{sin^{5}x cos | vedclass

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

Question

(B) माना $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} \right)

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}(8-\sqrt{6})$

Vedclass · Answer

(B) माना $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} ight) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} ight) dx$. $\int \frac{\sec^{2}x}{\sin^{5}x} dx$ पर खंडशः समाकलन (Integration by Parts) का उपयोग करने पर,$u = \frac{1}{\sin^{5}x}$ और $dv = \sec^{2}x dx$ लें। तब $du = -5 \sin^{-6}x \cos x dx$ और $v = 	an x$. $\int \frac{\sec^{2}x}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} - \int 	an x (-5 \sin^{-6}x \cos x) dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{\sin x / \cos x \cdot \cos x}{\sin^{6}x} dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx$. इस मान को $I$ में प्रतिस्थापित करने पर: $I = \left( \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx ight) - 5 \int \frac{1}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} + C$. अतः,$f(x) = \frac{	an x}{\sin^{5}x} = \frac{1}{\cos x \sin^{4}x}$. $f(\frac{\pi}{6}) = \frac{1}{\cos(\pi/6) \sin^{4}(\pi/6)} = \frac{32}{\sqrt{3}}$. $f(\frac{\pi}{4}) = \frac{1}{\cos(\pi/4) \sin^{4}(\pi/4)} = 4\sqrt{2}$. $f(\frac{\pi}{6}) - f(\frac{\pi}{4}) = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$.

यदि $\int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = f(x) + C$ जहाँ $C$ समाकलन स्थिरांक है,तो $f(\frac{\pi}{6}) - f(\frac{\pi}{4})$ का मान ज्ञात कीजिए।

Similar Questions

$\text{यदि } \int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{e}\left(4 e^{x}+7 e^{-x}\right)\right)+C$ जहाँ $C$ एक समाकलन स्थिरांक है,तो $u+v$ का मान .... है।

यदि $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ है,तो $\frac{\lambda}{\mu}=$

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।

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

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=A \cos x+B \log |f(x)|+c$ (जहाँ $c$ समाकलन का एक स्थिरांक है)। तो $A, B$ और $f(x)$ के मान हैं:

Vedclass Test Series

Exam Paper Generator

Online Exam Module