int frac{x^{2}}{(x sin x+cos x)^{2}} d x is e | vedclass

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

Question

(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$. We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$. We

Vedclass · Accepted Answer

$\frac{\sin x-x \cos x}{x \sin x+\cos x}+C$

Vedclass · Answer

(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$. We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$. We can rewrite the integral as: $I = \int \left( \frac{x}{\cos x} ight) \left( \frac{x \cos x}{(x \sin x+\cos x)^{2}} ight) d x$. Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x$. Then $du = \frac{\cos x - x(-\sin x)}{\cos^{2} x} d x = \frac{\cos x + x \sin x}{\cos^{2} x} d x$ and $v = \frac{-1}{x \sin x + \cos x}$. $I = \left( \frac{x}{\cos x} ight) \left( \frac{-1}{x \sin x + \cos x} ight) - \int \left( \frac{\cos x + x \sin x}{\cos^{2} x} ight) \left( \frac{-1}{x \sin x + \cos x} ight) d x$. $I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \frac{1}{\cos^{2} x} d x$. $I = \frac{-x}{\cos x(x \sin x + \cos x)} + 	an x + C$. $I = \frac{-x + 	an x \cos x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x + \sin x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x + x \sin^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x(1 - \sin^{2} x) + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{-x \cos^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{\cos x(\sin x - x \cos x)}{\cos x(x \sin x + \cos x)} + C$. $I = \frac{\sin x - x \cos x}{x \sin x + \cos x} + C$.

$\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$ is equal to

Similar Questions

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$, then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

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

Integrate the function: $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$

If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration,then the function $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module