If $y(x)=\int_{\pi^{2} / 16}^{x^{2}} \frac{\cos x \cdot \cos \sqrt{ heta}}{1+\sin ^{2} \sqrt{ heta

Question

If $y(x)=\int_{\pi^{2} / 16}^{x^{2}} \frac{\cos x \cdot \cos \sqrt{	heta}}{1+\sin ^{2} \sqrt{	heta}} d 	heta$, find $\frac{d y}{d x}$ at $x=\pi,$ is $n\pi$, find $n$

Professor PC Thomas Classes · Answer

$y=\int_{\pi^{2} / 16}^{x^{2}} \frac{\cos x \cdot \cos \sqrt{	heta}}{1+\sin ^{2} \sqrt{	heta}} d 	heta$\ $\Rightarrow y=\cos x \cdot \int_{\pi^{2} / 16}^{x^{2}} \frac{\cos \sqrt{	heta}}{1+\sin ^{2} \sqrt{	heta}} d 	heta$\ so that $\frac{d y}{d x}=-\sin x \int_{\pi^{2} / 16}^{x^{2}} \frac{\cos \sqrt{	heta}}{1+\sin ^{2} \sqrt{	heta}} d 	heta$ $$ +\frac{2 x \cos x \cdot \cos x}{1+\sin ^{2} x} $$ Hence, at $x=\pi, \frac{d y}{d x}=0+\frac{2 \pi(-1)(-1)}{1+0}=2 \pi$.\