Question

Let $x=\underset{0}{\overset{\cos t}{\int }}(z^2-1){\cos }^{2}zdz$ and $y=\underset{0}{\overset{{\sin }^{2}t}{\int }}{z}^2\left({\sin }^2\sqrt{1-z}\right)dz,t\in (0,\frac{\pi }{2}),$ then $\frac{dy}{dx}$ is equal to

• A. ${\sin }^{2}2t\cdot {\tan }^{2}t$
• B. $\sin t\cdot \sin 2t\cdot {\tan }^2\left(\cos t\right)$
• C. $\cos t\cdot \sin 2t\cdot \tan \left(\cos t\right)\cdot \cot \left(\cos t\right)$
• D. ${\sin }^{2}t\cdot \sin 2t\cdot {\cot }^2\left(\cos t\right)$

Answer 1

The correct option is B $\sin t\cdot \sin 2t\cdot {\tan }^2\left(\cos t\right)$

Given : $x=\underset{0}{\overset{\cos t}{\int }}(z^2-1){\cos }^{2}zdz$
differentiating both sides w.r.t. $t,$
$\Rightarrow \frac{dx}{dt}=({\cos }^{2}z-1){\cos }^2\left(\cos t\right)\cdot (-\sin t)\Rightarrow \frac{dx}{dt}={\sin }^{3}t\cdot {\cos }^2\left(\cos t\right)$
and $y=\underset{0}{\overset{{\sin }^{2}t}{\int }}{z}^2\left({\sin }^2\sqrt{(1-z)}\right)dz$
differentiating both sides w.r.t. $t,$
$\Rightarrow \frac{dy}{dt}={\sin }^{4}t\cdot \left({\sin }^2\sqrt{1-{\sin }^{2}t}\right)\cdot \left(2\sin t\cos t\right)\Rightarrow \frac{dy}{dt}=2{\sin }^{5}t\cos t{\sin }^2\left(\cos t\right)$
So, $\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\sin t\cdot \sin 2t\cdot {\tan }^2\left(\cos t\right)$