Chat with tutor

Ask Questions, Get Answers

Questions  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

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$

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$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.