Question

If $g\left( x \right)=\int\limits_{\sin \left( x \right)}^{\sin \left( 2x \right)}{{{\sin }^{-1}}\left( t \right)dt}$ , then
(a) ${g}'\left( \dfrac{\pi }{2} \right)=-2\pi$
(b) ${g}'\left( -\dfrac{\pi }{2} \right)=-2\pi$
(c) ${g}'\left( -\dfrac{\pi }{2} \right)=2\pi$
(d) ${g}'\left( \dfrac{\pi }{2} \right)=2\pi$
This question can have multiple correct answers.

Answer 1

We will use the formula ${f}'\left( x \right)=z\left( h\left( x \right) \right)\cdot {h}'\left( x \right)-z\left( g\left( x \right) \right)\cdot {g}'\left( x \right)$ for $f\left( x \right)=\int\limits_{g\left( x \right)}^{h\left( x \right)}{z\left( t \right)dt}$ to solve the integral given in the question. Then we will substitute the value of x by $\dfrac{\pi }{2}$ and $-\dfrac{\pi }{2}$ to find the correct answer (s) of the question.

Complete step-by-step answer:
We know for a function given in terms of an integral $f\left( x \right)=\int\limits_{g\left( x \right)}^{h\left( x \right)}{z\left( t \right)dt}$, its differentiation is given by ${f}'\left( x \right)=z\left( h\left( x \right) \right)\cdot {h}'\left( x \right)-z\left( g\left( x \right) \right)\cdot {g}'\left( x \right)$

We should recall the identity ${{\sin }^{-1}}\left( \sin \theta \right)=\theta ,\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\text{ }\ldots \left( i \right)$

Let us also recall a few values on trigonometric functions,
$\begin{aligned} & \cos \left( \dfrac{-\pi }{2} \right)=\cos \left( \dfrac{\pi }{2} \right)=0 \\\ & \cos \left( -\pi \right)=\cos \left( -\pi \right)=-1 \\\ & \sin \left( -\pi \right)=\sin \left( \pi \right)=0 \\\ \end{aligned}$

We are given $g\left( x \right)=\int\limits_{\sin \left( x \right)}^{\sin \left( 2x \right)}{{{\sin }^{-1}}\left( t \right)dt}$
Then, differentiating both sides of the above equation, we get

& {g}'\left( x \right)={{\sin }^{-1}}\left( \sin \left( 2x \right) \right)\cdot 2\cos \left( 2x \right)-{{\sin }^{-1}}\left( \sin \left( x \right) \right)\cdot \cos \left( x \right) \\\ & =2x\cdot 2\cos \left( 2x \right)-x\cdot \cos \left( x \right)\text{ from }\left( i \right) \\\ & {g}'\left( x \right)=4x\cos \left( 2x \right)-x\cos x\text{ }\ldots \left( ii \right) \end{aligned}$$ Now putting the value of $$x=\dfrac{-\pi }{2}$$ in equation (ii), we get $$\begin{aligned} & {g}'\left( \dfrac{-\pi }{2} \right)=4\left( \dfrac{-\pi }{2} \right)\cos \left( 2\dfrac{-\pi }{2} \right)-\dfrac{-\pi }{2}\cos \dfrac{-\pi }{2} \\\ & =-2\pi \cos \left( -\pi \right)+\dfrac{\pi }{2}\cos \left( \dfrac{-\pi }{2} \right) \\\ & =-2\pi \left( -1 \right)+0 \\\ & {g}'\left( \dfrac{-\pi }{2} \right)=2\pi \end{aligned}$$ Hence, the option (c) is correct. Now putting the value of $$x=\dfrac{\pi }{2}$$ in equation (ii), we get $$\begin{aligned} & {g}'\left( \dfrac{\pi }{2} \right)=4\left( \dfrac{\pi }{2} \right)\cos \left( 2\dfrac{\pi }{2} \right)-\dfrac{\pi }{2}\cos \dfrac{\pi }{2} \\\ & =2\pi \cos \left( \pi \right)+\dfrac{\pi }{2}\cos \left( \dfrac{\pi }{2} \right) \\\ & =2\pi \left( -1 \right)+0 \\\ & {g}'\left( \dfrac{\pi }{2} \right)=-2\pi \end{aligned}$$ Hence, the option (a) is correct. **So, the correct answer is “Option A and C”.** **Note:** The value of $${{\sin }^{-1}}\left( \sin \theta \right)=\theta ,\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$$ is the principal value of the identity. The question also doesn’t mention that the interval to be considered is $$\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$$. This was solved on the instinct by looking at the options wisely. Here, we are assuming that the integral is to be solved only in the interval $$\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$$. However, if we took general values of the identity, then the values of $${g}'\left( \dfrac{\pi }{2} \right)$$ and $${g}'\left( \dfrac{-\pi }{2} \right)$$ would have been zero. Let us look at the differentiation of the given integral $${g}'\left( x \right)={{\sin }^{-1}}\left( \sin \left( 2x \right) \right)\cdot 2\cos \left( 2x \right)-{{\sin }^{-1}}\left( \sin \left( x \right) \right)\cdot \cos \left( x \right)$$. Then, for $$x=\dfrac{-\pi }{2}$$, we get $$\begin{aligned} & {g}'\left( \dfrac{-\pi }{2} \right)={{\sin }^{-1}}\left( \sin \left( 2\dfrac{-\pi }{2} \right) \right)\cdot 2\cos \left( 2\dfrac{-\pi }{2} \right)-{{\sin }^{-1}}\left( \sin \left( \dfrac{-\pi }{2} \right) \right)\cdot \cos \left( \dfrac{-\pi }{2} \right) \\\ & ={{\sin }^{-1}}\left( \sin \left( -\pi \right) \right)\cdot 2\cos \left( -\pi \right)-0 \\\ & ={{\sin }^{-1}}0=0 \end{aligned}$$ And, for $$x=\dfrac{\pi }{2}$$, we get $$\begin{aligned} & {g}'\left( \dfrac{\pi }{2} \right)={{\sin }^{-1}}\left( \sin \left( 2\dfrac{\pi }{2} \right) \right)\cdot 2\cos \left( 2\dfrac{\pi }{2} \right)-{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right)\cdot \cos \left( \dfrac{\pi }{2} \right) \\\ & ={{\sin }^{-1}}\left( \sin \left( \pi \right) \right)\cdot 2\cos \left( \pi \right)-0 \\\ & ={{\sin }^{-1}}0=0 \end{aligned}$$ Hence, the actual general values of $${g}'\left( \dfrac{\pi }{2} \right)$$ and $${g}'\left( \dfrac{-\pi }{2} \right)$$ is zero. But since the options mentioned other values, we found out the principal values of the function.