If int frac{dx}{(x tan x + 1)^2} = f(x) + c,t | vedclass

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(D) We are given the integral $I = \int \frac{dx}{(x 	an x + 1)^2}$.Multiply the numerator and denominator by $\cos^2 x$:$I = \int \frac{\cos^2 x}{(\

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$\infty$

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(D) We are given the integral $I = \int \frac{dx}{(x 	an x + 1)^2}$.Multiply the numerator and denominator by $\cos^2 x$:$I = \int \frac{\cos^2 x}{(\cos x + x \sin x)^2} dx$.Let $u = \cos x + x \sin x$. Then $du = (-\sin x + \sin x + x \cos x) dx = x \cos x dx$.This does not simplify directly. Let us rewrite the integrand as $\int \frac{\cos^2 x}{(\cos x + x \sin x)^2} dx$.Note that $\frac{d}{dx} \left( \frac{x \sin x}{\cos x + x \sin x} ight) = \frac{(\sin x + x \cos x)(\cos x + x \sin x) - x \sin x (-\sin x + \sin x + x \cos x)}{(\cos x + x \sin x)^2} = \frac{\cos^2 x + x \sin x \cos x + x \sin x \cos x + x^2 \sin^2 x - x^2 \sin x \cos x}{(\cos x + x \sin x)^2} = \frac{\cos^2 x + x \sin x \cos x}{(\cos x + x \sin x)^2} = \frac{\cos x}{\cos x + x \sin x}$.Actually,consider $f(x) = \frac{-x \sin x}{\cos x + x \sin x}$.Then $f'(x) = \frac{-( \sin x + x \cos x)(\cos x + x \sin x) + x \sin x (x \cos x)}{(\cos x + x \sin x)^2} = \frac{-\sin x \cos x - x \sin^2 x - x \cos^2 x - x^2 \sin x \cos x + x^2 \sin x \cos x}{(\cos x + x \sin x)^2} = \frac{-\sin x \cos x - x}{(\cos x + x \sin x)^2}$.This suggests $f(x) = \frac{x \cos x}{\cos x + x \sin x}$.As $x ightarrow \frac{\pi}{2}$,$f(x) = \frac{x \cos x}{\cos x + x \sin x} = \frac{x}{\frac{\cos x}{\cos x} + x \frac{\sin x}{\cos x}} = \frac{x}{1 + x 	an x}$.As $x ightarrow \frac{\pi}{2}$,$f(x) ightarrow \frac{\pi/2}{1 + \infty} = 0$.

If $\int \frac{dx}{(x \tan x + 1)^2} = f(x) + c$,then $\lim_{x \rightarrow \frac{\pi}{2}} f(x) = $

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