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$I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} = ?$

$I = \int\limits_0^\pi  {\frac{{2(\pi  - x)\sin (\pi  - x)}}{{3 + \cos 2(\pi  - x)}}dx} $

$ = \int\limits_0^\pi  {\frac{{2(\pi  - x)\sin x}}{{3 + \cos 2x}}dx} $

$2I = \int\limits_0^\pi  {\frac{{2x\sin x}}{{3 + \cos 2x}}dx}  + \int\limits_0^\pi  {\frac{{2(\pi  - x)\sin x}}{{3 + \cos 2x}}dx}  = \int\limits_0^\pi  {\frac{{2\pi \sin x}}{{3 + \cos 2x}}dx} $

$ \Rightarrow I = \pi \int\limits_0^\pi  {\frac{{\sin x}}{{3 + \cos 2x}}dx} $

$ = \pi \int\limits_0^\pi  {\frac{{\sin x}}{{3 + 2{{\cos }^2}x - 1}}dx} $

$ = \frac{\pi }{2}\int\limits_0^\pi  {\frac{{\sin x}}{{1 + {{\cos }^2}x}}dx} $

Let, $\cos x = t$

$ \Rightarrow  - \sin xdx = dt$

$I = \frac{\pi }{2}\int\limits_1^{ - 1} {\frac{{ - dt}}{{1 + {t^2}}}} $

$ = \frac{\pi }{2}\int\limits_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \frac{{\pi  \times 2}}{2}\int\limits_0^1 {\frac{{dt}}{{1 + {t^2}}}} $

$ = \pi \left. {{{\tan }^{ - 1}}t} \right|_0^1$

$ = \frac{{{\pi ^2}}}{4}$

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$