The integral intleft(frac{x}{x sin x+cos x}ri | vedclass

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(D) Let $I = \int \left(\frac{x}{x \sin x + \cos x}\right)^2 dx$. We can rewrite the integrand as: $I = \int \left(\frac{x \sec x}{x \sin x + \cos x}\

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \left(\frac{x}{x \sin x + \cos x}ight)^2 dx$. We can rewrite the integrand as: $I = \int \left(\frac{x \sec x}{x \sin x + \cos x}ight) \cdot \left(\frac{\cos x}{x \sin x + \cos x}ight) dx$. Using integration by parts,let $u = x \sec x$ and $dv = \frac{\cos x}{(x \sin x + \cos x)^2} dx$. Then $du = (\sec x + x \sec x 	an x) dx = \frac{\cos x + x \sin x}{\cos^2 x} dx$ and $v = -\frac{1}{x \sin x + \cos x}$. Applying the formula $\int u dv = uv - \int v du$: $I = (x \sec x) \left(-\frac{1}{x \sin x + \cos x}ight) - \int \left(-\frac{1}{x \sin x + \cos x}ight) \left(\frac{\cos x + x \sin x}{\cos^2 x}ight) dx$. $I = -\frac{x \sec x}{x \sin x + \cos x} + \int \frac{1}{\cos^2 x} dx$. $I = -\frac{x \sec x}{x \sin x + \cos x} + \int \sec^2 x dx$. $I = 	an x - \frac{x \sec x}{x \sin x + \cos x} + C$.

The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to (where $C$ is a constant of integration)

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