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If $y(x) = \Large \int \limits_{\frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{\cos x . \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$, find $\;y'(\pi)$

(A) $\large\frac{\pi^2}{8}$ $\quad$ (B) $\large\frac{\pi^2}{16}$ $\quad$ (C) $2 \pi$ $\quad$ (D) $\large\frac{\pi^2}{16}$$ - 2 \pi$

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1 Answer

We can immediately see that we are integrating with respect to $\theta \rightarrow$ $x$ and any expression depending on $x$ is a constant.
$\Rightarrow y(x) = \cos x \Large \int \limits_{\normalsize \frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{ \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$
$y(x)$ is a product of two expressions involving $x$. For the second expression, which is the integral, we can now apply the fundamental theorem as we have removed the dependence of $x$ outside the integrand.
$\Rightarrow y'(x) = -\sin x \Large \int \limits_{\normalsize \frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{ \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$$ + \cos x \large\frac{d}{dx} \Large \int \limits_{\normalsize \frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{ \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$
$\Rightarrow y'(x) = -\sin x \Large \int \limits_{\normalsize \frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{ \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$$ + \cos x \large\frac{d(x^2)}{dx} \large\frac{ \cos \sqrt x^2 }{1+\sin^2 \sqrt x^2}$
$\Rightarrow y'(x) = -\sin x \Large \int \limits_{\normalsize \frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{ \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$$ + \large\frac{2x\cos^2 x}{1+\sin^2x}$
We are solving for $y'(\pi)$. Since $\sin \pi = 0$, the first term $ = 0$.
The second term reduces to $y'(\pi) = \large\frac{2\pi \cos^2 \pi}{1+\sin^2 \pi}$$ = 2 \pi$
answered Dec 23, 2013 by meena.p
edited Mar 26, 2014 by balaji

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