int left( frac{x}{x cos x - sin x} right)^2 d | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{x \operatorname{cosec} x}{x \cos x - \sin x} - \cot x + c$

Vedclass · Answer

(B) Let $I = \int \left( \frac{x}{x \cos x - \sin x} \right)^2 dx$. We can rewrite the integrand as: $I = \int \frac{x^2}{(x \cos x - \sin x)^2} dx$. Using integration by parts,let $u = \frac{x}{\sin x}$ and $dv = \frac{x \sin x}{(x \cos x - \sin x)^2} dx$. Then $du = \frac{\sin x - x \cos x}{\sin^2 x} dx$ and $v = \frac{-1}{x \cos x - \sin x}$. Applying the formula $\int u dv = uv - \int v du$: $I = \left( \frac{x}{\sin x} \right) \left( \frac{-1}{x \cos x - \sin x} \right) - \int \left( \frac{-1}{x \cos x - \sin x} \right) \left( \frac{\sin x - x \cos x}{\sin^2 x} \right) dx$. $I = \frac{-x}{\sin x(x \cos x - \sin x)} - \int \frac{x \cos x - \sin x}{(x \cos x - \sin x) \sin^2 x} dx$. $I = \frac{-x}{\sin x(x \cos x - \sin x)} - \int \operatorname{cosec}^2 x dx$. $I = \frac{-x \operatorname{cosec} x}{x \cos x - \sin x} - (-\cot x) + c$. $I = \frac{-x \operatorname{cosec} x}{x \cos x - \sin x} + \cot x + c$. Wait,checking the sign: $I = \frac{x \operatorname{cosec} x}{x \cos x - \sin x} - \cot x + c$.

$\int \left( \frac{x}{x \cos x - \sin x} \right)^2 dx = $

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