Question

The value of $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt$ is:
A) $\dfrac{\pi }{2}$
B) $1$
C) $\dfrac{\pi }{4}$
D) None of these

Answer 1

Here we have to find the value of $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt$. Take and solve first part i.e. $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt$. In the first part assume $t={{\sin }^{2}}\theta$ and solve it. Also, for the second part assume $t={{\cos }^{2}}\theta$. After simplifying them apply some properties and add them. Try it, you will definitely get the answer.

Complete step by step solution:
Here we are given $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt$.
So to solve this let us take first part i.e. $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt$.
In this take $t={{\sin }^{2}}\theta$ .
Now differentiating $t={{\sin }^{2}}\theta$ we get,
$dt=2\sin \theta \cos \theta d\theta =\sin 2\theta d\theta$
So when $t=0$ then $\theta =0$. Also, when $t={{\sin }^{2}}x$ then $\theta =x$.
So substituting all values in first part we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{{{\sin }^{-1}}(\sqrt{{{\sin }^{2}}\theta })}\sin 2\theta d\theta$
Now simplifying we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{{{\sin }^{-1}}(\sqrt{{{\sin }^{2}}\theta })}\sin 2\theta d\theta$
Again, simplifying in simple manner we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{{{\sin }^{-1}}(\sin \theta )}\sin 2\theta d\theta$
Again, simplifying we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{\theta }\sin 2\theta d\theta$ ……….. (1)
Now let us take second part i.e. $\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt$.
In this take $t={{\cos }^{2}}\theta$ .
Now differentiating $t={{\cos }^{2}}\theta$ we get,
$dt=-2\sin \theta \cos \theta d\theta =-\sin 2\theta d\theta$
So, when $t=0$ then $\theta =\dfrac{\pi }{2}$. Also, when $t={{\cos }^{2}}x$ then $\theta =x$.
Substituting all values in first part we get,
$\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\int\limits_{\dfrac{\pi }{2}}^{x}{{{\cos }^{-1}}(\sqrt{{{\cos }^{2}}\theta })}(-\sin 2\theta )d\theta$
Now simplifying we get,
$\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\int\limits_{\dfrac{\pi }{2}}^{x}{{{\cos }^{-1}}({{\cos }^{2}}\theta )}(-\sin 2\theta )d\theta$
$\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=-\int\limits_{\dfrac{\pi }{2}}^{x}{\theta }\sin 2\theta d\theta$
So, we know the property that, if there is minus sign so if we want to remove it the limits will interchange. So, it becomes
$\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\int\limits_{x}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta$ …….. (2)
Now adding equation (1) and (2) we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{\theta }\sin 2\theta d\theta +\int\limits_{x}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta$
Now we know the property that, $\int\limits_{a}^{b}{f(x)dx+}\int\limits_{b}^{c}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx}$.
Applying above property we get,
$\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\int\limits_{0}^{x}{\theta }\sin 2\theta d\theta +\int\limits_{x}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta$
So now taking i.e. integrating $\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta$.
Now integrating the above by Integration by parts we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\left[ \theta \times \dfrac{(-\cos 2\theta )}{2} \right]_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{(-\cos 2\theta )}{2}d\theta }$
Now simplifying we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\left[ \theta \times \dfrac{(-\cos 2\theta )}{2} \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2\theta d\theta }$ ……….. (3)
Now let us find the integration of $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2\theta d\theta }$,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2\theta d\theta }=\left[ \dfrac{\sin 2\theta }{2} \right]_{0}^{\dfrac{\pi }{2}}$
Now substituting $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2\theta d\theta }=\left[ \dfrac{\sin 2\theta }{2} \right]_{0}^{\dfrac{\pi }{2}}$ in equation (3) we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\left[ \theta \times \dfrac{(-\cos 2\theta )}{2} \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \dfrac{\sin 2\theta }{2} \right]_{0}^{\dfrac{\pi }{2}}$
Now applying limit we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\left[ \theta \times \dfrac{(-\cos 2\theta )}{2} \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \dfrac{\sin 2\theta }{2} \right]_{0}^{\dfrac{\pi }{2}}=\left[ \dfrac{\pi }{2}\times \left( -\dfrac{1}{2} \right)\times \cos \pi \right]$
Here, $\cos \pi =-1$. Simplifying we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\left[ \dfrac{\pi }{2}\times \left( -\dfrac{1}{2} \right)\times -1 \right]$
$\int\limits_{0}^{\dfrac{\pi }{2}}{\theta }\sin 2\theta d\theta =\dfrac{\pi }{4}$
Therefore, $\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}}dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt=\dfrac{\pi }{4}$.

We get the correct answer as option (C).

Note:
Here in the problem, you must know the assumption as we have took $t={{\sin }^{2}}\theta$ in first part and $t={{\cos }^{2}}\theta$ in second part. The properties related to minus sign and integration must be known. While substitution be aware that no terms are missing.