text{If } int frac{1}{operatorname{cosec} x+c | vedclass

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

Question

(A) $	ext{Given } I = \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \int \frac{\sin x}{1+\sin x \cos x} d x = \int \frac{2 \sin x}{2+\sin 2 x} d

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}-1}{\sqrt{3}+1}$

Vedclass · Answer

(A) $	ext{Given } I = \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \int \frac{\sin x}{1+\sin x \cos x} d x = \int \frac{2 \sin x}{2+\sin 2 x} d x$ $	ext{We can write } 2 \sin x = (\sin x + \cos x) - (\cos x - \sin x)$ $	ext{So, } I = \int \frac{\sin x + \cos x}{2 + \sin 2x} d x - \int \frac{\cos x - \sin x}{2 + \sin 2x} d x$ $	ext{Comparing with the given expression, } \frac{1}{2 \sqrt{3}} \log |f(x)| = \int \frac{\sin x + \cos x}{2 + \sin 2x} d x$ $	ext{Note that } 2 + \sin 2x = 3 - (1 - \sin 2x) = 3 - (\sin x - \cos x)^2$ $	ext{Let } u = \sin x - \cos x, 	ext{ then } du = (\cos x + \sin x) d x$ $\int \frac{du}{3 - u^2} = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + u}{\sqrt{3} - u} ight| + C$ $	ext{Thus, } f(x) = \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x}$ $	ext{At } x = \frac{\pi}{3}, \sin x = \frac{\sqrt{3}}{2} 	ext{ and } \cos x = \frac{1}{2}$ $|f(\frac{\pi}{3})| = \left| \frac{\sqrt{3} + \frac{\sqrt{3}}{2} - \frac{1}{2}}{\sqrt{3} - \frac{\sqrt{3}}{2} + \frac{1}{2}} ight| = \frac{\frac{3 \sqrt{3} - 1}{2}}{\frac{\sqrt{3} + 1}{2}} = \frac{3 \sqrt{3} - 1}{\sqrt{3} + 1}$

$\text{If } \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \frac{1}{2 \sqrt{3}} \log |f(x)| - \int \frac{\cos x-\sin x}{2+\sin 2 x} d x + c, \text{ then at } x = \frac{\pi}{3}, |f(x)| = $

Similar Questions

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ is equal to

Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$ and $J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$. Then,for an arbitrary constant $C$,the value of $J-I$ equals

$\int \frac{3 \sin x+5 \cos x+4}{\sin x+\cos x+2} d x=$

$\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx = $

$\int \frac{d x}{(x+1) \sqrt{x^2+4}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module