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In the below integral we have to find the value of f'(x)
\[f\left( x \right)=\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt}\]
\[\begin{align}
  & \left( a \right)\dfrac{\pi }{2} \\
 & \left( b \right)1 \\
 & \left( c \right)\dfrac{\pi }{4} \\
 & \left( d \right)0 \\
\end{align}\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: In this question, we first need to find the integration of the given respective terms within the proper limits and then find the value of the function that would be obtained after the integration. Then find the derivative of the function so obtained to get the answer.

Complete step-by-step answer:
Let f(x) is a function differentiable in an interval [a, b]. i.e. at every point of the interval, the derivative of the function exists finitely and is unique.
Hence, we may define new function g(x) = f'(x) called the differentiation of the function f(x) with respect to x and is denoted by:
\[f'\left( x \right)=\dfrac{d}{dx}f\left( x \right)\]
INTEGRATION: Let f(x) be a function. Then, the collection of all its primitives is called the indefinite integral or anti-derivative of f(x) and is denoted by:
\[\int{f\left( x \right)dx}\]
Integration is an inverse process of differentiation.
The process of finding functions whose derivative is given, is called anti-differentiation or integration.
\[\dfrac{d}{dx}\left\{ \varphi \left( x \right) \right\}=f\left( x \right)\Leftrightarrow \int{f\left( x \right)dx}=\varphi \left( x \right)+C\]
Where, C is called the constant of integration.
DEFINITE INTEGRAL: Let f(x) be a function defined on the interval [a, b] and F(x) be its antiderivative. Then,
\[\int\limits_{a}^{b}{f\left( x \right)dx=F\left( b \right)-F\left( a \right)}\]
is defined as the definite integral of f(x) from x=a to x=b.
The numbers a and b are called upper and lower limits of integration.
Now, let us consider the first part of the given function.
\[\Rightarrow \int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}\]
Let us assume that:
\[\sqrt{t}=u\]
Now, by differentiation on both sides we get,
\[\begin{align}
  & \dfrac{1}{2}\cdot \dfrac{1}{\sqrt{t}}dt=du \\
 & dt=2udu\text{ }\left[ \because \sqrt{t}=u \right] \\
\end{align}\]
Thus, the limit changes in the following way:
Then the lower limit of the integral is 0.
And the upper limit is calculated as follows:
Now, we know that in the integral given in “t” in the above question,
\[\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}\]
The lower limit and upper limit are individually equal to t.
Then in the below, we have shown how lower and upper limit is written in the definite integral:
$t=0$
$t={{\sin }^{2}}x$
Now, we have assumed that, \[\sqrt{t}=u\] and we are going to completely convert the given integral in “t” to “u” so taking square root on both the sides we get,
$\begin{align}
  & \sqrt{t}=\sqrt{{{\sin }^{2}}x} \\
 & \Rightarrow \sqrt{t}=\sin x \\
\end{align}$
And as we have assumed that \[\sqrt{t}=u\] so putting \[\sqrt{t}=u\] in the above equation we get the upper limit of the definite integral in “u” is:
$u=\sin x$
Hence, we got the lower limit and upper limit of the new definite integral in “u is 0 and $\sin x$ respectively.
By substituting these in the above equation we get,
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{\sin x}{{{\sin }^{-1}}u\cdot 2udu} \\
 & \Rightarrow 2\int\limits_{0}^{\sin x}{u{{\sin }^{-1}}udu} \\
\end{align}\]
Now, the above integration has two functions in u which are in multiplication with each other so this kind of integration is done using integration by parts which we are showing as follows:
\[\begin{align}
  & \int{u{{\sin }^{-1}}u}du={{\sin }^{-1}}u\int{udu}-\int{\dfrac{d}{du}\left( {{\sin }^{-1}}u \right)}\cdot \int{udu} \\
 & \left[ \because \int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\dfrac{d}{dx}\left( f\left( x \right) \right)}\cdot \int{g\left( x \right)dx}} \right] \\
\end{align}\]
As we already know that on solving the above equation we get ,
\[\int{u{{\sin }^{-1}}u}du={{\sin }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)-\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\]……… Eq. (1)
Now, we are showing the integration of \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\]:
Let us assume $u=\sin \theta $ in the above integral.
Then differentiating both the sides of this assumption we get,
$du=\cos \theta d\theta $
Substituting these values of u and du in \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\] we get,
\[\begin{align}
  & \int{\dfrac{1}{\sqrt{1-{{\sin }^{2}}\theta }}\left( \dfrac{{{\sin }^{2}}\theta }{2} \right)}\cos \theta d\theta \\
 & =\int{\dfrac{1}{\cos \theta }}\left( \dfrac{{{\sin }^{2}}\theta }{2} \right)\cos \theta d\theta \\
 & =\int{\dfrac{{{\sin }^{2}}\theta }{2}}d\theta \\
\end{align}\]
We can write ${{\sin }^{2}}\theta $ as:
$\dfrac{1-\cos 2\theta }{2}$
\[\begin{align}
  & \int{\dfrac{1}{2}}\left( \dfrac{1-\cos 2\theta }{2} \right)d\theta \\
 & =\dfrac{1}{4}\int{\left( 1-\cos 2\theta \right)}d\theta \\
 & =\dfrac{1}{4}\left( \int{d\theta }-\int{\cos 2\theta d\theta } \right) \\
 & =\dfrac{1}{4}\left( \theta -\dfrac{1}{2}\sin 2\theta \right) \\
 & =\dfrac{1}{4}\left( \theta -\dfrac{1}{2}2\sin \theta \cos \theta \right) \\
 & =\dfrac{1}{4}\left( \theta -\sin \theta \cos \theta \right) \\
\end{align}\]
Now, substituting $\theta ={{\sin }^{-1}}u$ because we have assumed $u=\sin \theta $.
\[\dfrac{1}{4}\left( {{\sin }^{-1}}u-\sin \left( {{\sin }^{-1}}u \right)\cos \left( {{\sin }^{-1}}u \right) \right)\]
\[\dfrac{1}{4}\left( {{\sin }^{-1}}u-u\cos \left( {{\sin }^{-1}}u \right) \right)\]………. Eq. (2)
Now, we are going to solve $\cos \left( {{\sin }^{-1}}u \right)$ by assuming ${{\sin }^{-1}}u$ as $\alpha $ we get,
${{\sin }^{-1}}u=\alpha $
Taking sine on both the sides we get,
$u=\sin \alpha $
What we have to find is $\cos \left( {{\sin }^{-1}}u \right)$ in which if we put ${{\sin }^{-1}}u$ as $\alpha $ we get $\cos \alpha $ and we know that:
$\cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha }$
Substituting $u=\sin \alpha $ in the above equation we get,
$\cos \alpha =\sqrt{1-{{u}^{2}}}$
Hence, we have calculated the value of $\cos \left( {{\sin }^{-1}}u \right)$ as $\sqrt{1-{{u}^{2}}}$.
Substituting this value in eq. (2) we get,
\[\dfrac{1}{4}\left( {{\sin }^{-1}}u-u\sqrt{1-{{u}^{2}}} \right)\]
Substituting this value in eq. (1) in place of \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\] we get,
\[\int{u{{\sin }^{-1}}u}du={{\sin }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)-\dfrac{1}{4}\left( {{\sin }^{-1}}u-u\sqrt{1-{{u}^{2}}} \right)+C\]
Now, applying this integral value in \[2\int\limits_{0}^{\sin x}{u{{\sin }^{-1}}udu}\] we get,
\[2\int\limits_{0}^{\sin x}{u{{\sin }^{-1}}udu}=2\left( {{\sin }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)-\dfrac{1}{4}{{\sin }^{-1}}u +\dfrac{1}{4}u\sqrt{1-{{u}^{2}}} \right)_{0}^{\sin x}\]
\[\left[ \because \int\limits_{a}^{b}{f\left( x \right)dx=F\left( b \right)-F\left( a \right)} \right]\]
\[\Rightarrow 2\left( \dfrac{{{\sin }^{2}}x}{2}\times {{\sin }^{-1}}\left( \sin x \right)-\dfrac{{{\sin }^{-1}}\left( \sin x \right)}{4}+\dfrac{\sin x\sqrt{1-{{\sin }^{2}}x}}{4}-0 \right)\]
\[\Rightarrow x{{\sin }^{2}}x-\dfrac{x}{2}+\dfrac{\cos x\cdot \sin x}{2}\]
\[\left[ \begin{align}
  & \because {{\sin }^{-1}}\left( \sin x \right)=x \\
 & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \sqrt{{{\cos }^{2}}x}=\cos x \\
\end{align} \right]\]
\[\left[ \because {{\sin }^{-1}}0=0 \right]\]
Let us now solve the second part of the given function.
\[\Rightarrow \int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt}\]
Similarly, we can integrate the above as we have done for the first part, the only difference is the upper limit as $\cos x$ and the function is ${{\cos }^{-1}}$.
Let us assume that:
\[\sqrt{t}=u\]
Thus, the limits changes as:
\[u=0\text{ to u}=\cos x\]
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{\cos x}{{{\cos }^{-1}}u\cdot 2udu} \\
 & \Rightarrow 2\int\limits_{0}^{\cos x}{u{{\cos }^{-1}}udu} \\
\end{align}\]
Now, the above integration has two functions in u which are in multiplication with each other so this kind of integration is done using integration by parts which we are showing as follows:
\[\begin{align}
  & \int{u{{\cos }^{-1}}u}du={{\cos }^{-1}}u\int{udu}-\int{\dfrac{d}{du}\left( {{\cos }^{-1}}u \right)}\cdot \int{udu} \\
 & \left[ \because \int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\dfrac{d}{dx}\left( f\left( x \right) \right)}\cdot \int{g\left( x \right)dx}} \right] \\
\end{align}\]
As we already know that on solving the above equation we get ,
\[\int{u{{\cos }^{-1}}u}du={{\cos }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)+\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\]……… Eq. (3)
Now, we are showing the integration of \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\]:
Let us assume $u=\cos \theta $ in the above integral.
Then differentiating both the sides of this assumption we get,
$du=-\sin \theta d\theta $
Substituting these values of u and du in \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\] we get,
\[\begin{align}
  & -\int{\dfrac{1}{\sqrt{1-{{\cos }^{2}}\theta }}\left( \dfrac{{{\cos }^{2}}\theta }{2} \right)}\sin \theta d\theta \\
 & =-\int{\dfrac{1}{\sin \theta }}\left( \dfrac{{{\cos }^{2}}\theta }{2} \right)\sin \theta d\theta \\
 & =-\int{\dfrac{{{\cos }^{2}}\theta }{2}}d\theta \\
\end{align}\]
We can write ${{\cos }^{2}}\theta $ as:
$\dfrac{1+\cos 2\theta }{2}$
\[\begin{align}
  & -\int{\dfrac{1}{2}}\left( \dfrac{1+\cos 2\theta }{2} \right)d\theta \\
 & =-\dfrac{1}{4}\int{\left( 1+\cos 2\theta \right)}d\theta \\
 & =-\dfrac{1}{4}\left( \int{d\theta }+\int{\cos 2\theta d\theta } \right) \\
 & =-\dfrac{1}{4}\left( \theta +\dfrac{1}{2}\sin 2\theta \right) \\
 & =-\dfrac{1}{4}\left( \theta +\dfrac{1}{2}2\sin \theta \cos \theta \right) \\
 & =-\dfrac{1}{4}\left( \theta +\sin \theta \cos \theta \right) \\
\end{align}\]
Now, substituting $\theta ={{\cos }^{-1}}u$ because we have assumed $u=\cos \theta $.
\[\dfrac{1}{4}\left( {{\cos }^{-1}}u+\sin \left( {{\cos }^{-1}}u \right)\cos \left( {{\cos }^{-1}}u \right) \right)\]
\[-\dfrac{1}{4}\left( {{\cos }^{-1}}u+u\sin \left( {{\cos }^{-1}}u \right) \right)\]………. Eq. (4)
Now, we are going to solve $\sin \left( {{\cos }^{-1}}u \right)$ by assuming ${{\cos }^{-1}}u$ as $\alpha $ we get,
${{\cos }^{-1}}u=\alpha $
Taking cosine on both the sides we get,
$u=\cos \alpha $
What we have to find is $\sin \left( {{\cos }^{-1}}u \right)$ in which if we put ${{\cos }^{-1}}u$ as $\alpha $ we get $\sin \alpha $ and we know that:
$\sin \alpha =\sqrt{1-{{\cos }^{2}}\alpha }$
Substituting $u=\cos \alpha $ in the above equation we get,
$\sin \alpha =\sqrt{1-{{u}^{2}}}$
Hence, we have calculated the value of $\sin \left( {{\cos }^{-1}}u \right)$ as $\sqrt{1-{{u}^{2}}}$.
Substituting this value in eq. (4) we get,
\[-\dfrac{1}{4}\left( {{\cos }^{-1}}u+u\sin \left( {{\cos }^{-1}}u \right) \right)\]
\[\Rightarrow -\dfrac{1}{4}\left( {{\cos }^{-1}}u+u\sqrt{1-{{u}^{2}}} \right)\]
Substituting this value in eq. (3) in place of \[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}\left( \dfrac{{{u}^{2}}}{2} \right)}du\] we get,
\[\int{u{{\cos }^{-1}}u}du={{\cos }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)-\dfrac{1}{4}\left( {{\cos }^{-1}}u+u\sqrt{1-{{u}^{2}}} \right)+C\]
Now, applying this integral value in \[2\int\limits_{0}^{\cos x}{u{{\cos }^{-1}}udu}\] we get,
\[2\int\limits_{0}^{\cos x}{u{{\cos }^{-1}}udu}=\left( {{\cos }^{-1}}u\left( \dfrac{{{u}^{2}}}{2} \right)-\dfrac{1}{4}\left( {{\cos }^{-1}}u+u\sqrt{1-{{u}^{2}}} \right)+C \right)_{0}^{\cos x}\]
Applying the lower and upper limits in the above equation we get,
\[\begin{align}
  & \Rightarrow 2{{\left( \dfrac{{{x}^{2}}}{2}{{\cos }^{-1}}x-\dfrac{1}{4}{{\cos }^{-1}}x-\dfrac{1}{4}x\sqrt{1-{{x}^{2}}} \right)}_{0}}^{\cos x} \\
 & \left[ \because \int\limits_{a}^{b}{f\left( x \right)dx=F\left( b \right)-F\left( a \right)} \right] \\
\end{align}\]
\[\Rightarrow 2\left( \dfrac{{{\cos }^{2}}x}{2}\times co{{s}^{-1}}\left( cosx \right)-\dfrac{co{{s}^{-1}}\left( cosx \right)}{4}-\dfrac{cosx\sqrt{1-{{\cos }^{2}}x}}{4}+\dfrac{\pi }{8} \right)\]
\[\Rightarrow xco{{s}^{2}}x-\dfrac{x}{2}-\dfrac{\cos x\cdot \sin x}{2}+\dfrac{\pi }{4}\]
\[\left[ \begin{align}
  & \because co{{s}^{-1}}\left( cosx \right)=x \\
 & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \sqrt{{{\sin }^{2}}x}=\sin x \\
\end{align} \right]\]
\[\left[ \because {{\cos }^{-1}}0=\dfrac{\pi }{2} \right]\]
Now, we get from the above solved parts as:
\[\begin{align}
&f\left( x \right)=x{{\sin }^{2}}x-\dfrac{x}{2}+\dfrac{1}{2}\cos x\sin x+x{{\cos }^{2}}x-\dfrac{x}{2}-\dfrac{1}{2}\cos x\sin x+\dfrac{\pi }{4} \\
 &\Rightarrow f\left( x \right)=x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-x+\dfrac{\pi }{4} \left[ \because {{\sin }^{2}}x+{{\cos }^{2}}x=1 \right] \\
 & \Rightarrow f\left( x \right)=\dfrac{\pi }{4} \\
\end{align}\]
Now, we need to find the value of:
\[\begin{align}
  & \Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( f\left( x \right) \right) \\
 &\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\pi }{4} \right) \\
 & \Rightarrow f'\left( x \right)=0\text{ }\left[ \because \dfrac{d}{dx}\left( \text{constant} \right)=0 \right] \\
\end{align}\]
Hence, the correct option is (d).

Note: While replacing the given terms with other terms in order to integrate we need to be clear about the differential function it is going to change and we also need to change the limits of the integration accordingly if not then the result is going to be completely wrong.
We need to calculate the integration properly even with the signs because neglecting any of the signs will be a huge mistake. Here, while applying the value of the limits in the obtained result after integration we need to check the respective values we are going to obtain on applying the inverse conditions.
In the end, we may get confused with option (c) because the value of the given function we got after applying the integration is in option (c). But, in the given question we were asked to find the derivative of the obtained function. Hence, the answer will be 0.