Interesting Integrals 10

Let $\begin{aligned} I & = \color{#3D99F6}{\int_0^{\sin^2 x} \sin^{-1} \left(\sqrt t \right) dt} + \color{#D61F06}{\int_0^{\cos^2 x} \cos^{-1} \left( \sqrt t \right) dt} = \color{#3D99F6}{I_1} + \color{#D61F06}{I_2} \end{aligned}$

Now consider $I_1$ :

$\begin{aligned} I_1 & = \int_0^{\sin^2 x} \sin^{-1} \left(\sqrt t \right) dt \quad \quad \small \color{#3D99F6}{\text{Let }u^2 = t \implies 2u \ du = dt} \\ & = \int_0^{\sin x} 2u \sin^{-1} (u) \ du \quad \quad \small \color{#3D99F6}{\text{By integration by parts.}} \\ & = 2u \left(u \sin^{-1}(u) + \sqrt{1-u^2} \right) \bigg|_0^{\sin x} - \int_0^{\sin x} \left(2u \sin^{-1}(u) + 2\sqrt{1-u^2} \right) du \\ & = 2x \sin^2 x + 2\sin x \cos x - I_1 - 2 \int_0^{\sin x} \sqrt{1-u^2} \ du \quad \quad \small \color{#3D99F6}{\text{Let }u = \sin \theta \implies du = \cos \theta \ d \theta} \\ \implies 2I_1 & = 2x \sin^2 x + \sin 2 x - 2 \int_0^x \cos^2 \theta \ d \theta \\ & = 2x \sin^2 x + \sin 2 x - \int_0^x \left(1 + \cos 2\theta \right) d \theta \\ & = 2x \sin^2 x + \sin 2 x - x - \frac 12 \sin 2x \\ \implies I_1 & = x \sin^2 x + \frac 14 \sin 2 x - \frac x2 \end{aligned}$

Similarly for $I_2$ :

$\begin{aligned} I_2 & = \int_0^{\cos^2 x} \cos^{-1} \left(\sqrt t \right) dt \\ & = \int_0^{\cos x} 2u \cos^{-1} (u) \ du \\ & = 2u \left(u \cos^{-1}(u) - \sqrt{1-u^2} \right) \bigg|_0^{\cos x} - \int_0^{\cos x} \left(2u \cos^{-1}(u) - 2\sqrt{1-u^2} \right) du \\ & = 2x \cos^2 x - 2\cos x \sin x - I_2 + 2 \int_0^{\cos x} \sqrt{1-u^2} \ du \\ \implies 2I_2 & = 2x \cos^2 x - \sin 2x + 2 \int_0^{\frac \pi 2 -x} \cos^2 \theta \ d \theta \\ & = 2x \cos^2 x - \sin 2 x + \int_0^{\frac \pi 2 -x} \left(1 + \cos 2\theta \right) d \theta \\ & = 2x \cos^2 x - \sin 2 x + \frac \pi 2 - x + \frac 12 \sin 2x \\ \implies I_2 & = x \cos^2 x - \frac 14 \sin 2 x + \frac \pi 4 - \frac x2 \end{aligned}$

Therefore, finally we have:

$\begin{aligned} \implies I & = I_1+I_2 \\ & = x \sin^2 x + \frac 14 \sin 2 x - \frac x2 + x \cos^2 x - \frac 14 \sin 2 x + \frac \pi 4 - \frac x2 \\ & = \frac \pi 4 \end{aligned}$

$\implies c = \boxed{4}$

Its a nice problem.

Differentiate function on left hand side wrt x using Newton Leibnitz theorem to get the derivative of the function zero.

which implies that the given function is constant function

so put x = pi/4 and evaluate

as simple as that :)

Use ${cos}^{-1}x = \pi/2 - {sin}^{-1}x$ and the expression simplifies to

$(\pi/2) (cosx)^2 +$ integration of $sin (\sqrt{t} )$ from $(cosx)^2$ to $(sinx)^2$

Then you can easily integrate the expression 1st by substituting $\sqrt{t} = u$ followed by taking ${sin}^{-1}u = p$ :)

$I = \int_0^{\sin^2 x} \sin^{-1} \left(\sqrt t \right) dt + \int_0^{\cos^2 x} \cos^{-1} \left( \sqrt t \right) dt$

Let $u=\sin^{-1} \left(\sqrt t \right)$ for the first integral and $u = \cos^{-1} \left(\sqrt t \right)$ for the second integral.

$dt = \sin{2u} du$ and $dt = -\sin{2u} du$ respectively.

$I = \int_0^x u\sin{2u} du - \int_{\frac{\pi}{2}}^x u\sin2u du$ $=\int_0^{\frac{\pi}{2}} u\sin{2u} du$

Using Integration by Parts we get

$\Big [-\frac12 u\cos{2u} \Big]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos{2u} du$ $=\frac{\pi}{4} + \Big [ \frac12 \sin{2u} \Big]_0^\frac{\pi}{2}$ $=\boxed{\dfrac{\pi}{4}}$

The question is JEE style and so is the method I provide. Since the answer is independent of x, I choose x=pi/4.Using the property, arc (sin y) +arc (cos y)=pi/2, for any suitable y in the domain,the answer is simply (1/2) times pi/2=pi/4.

Since c is a constant, the integral must not depend on the value of $x$ .

Let $x=\dfrac{\pi}{4}$

$I=\displaystyle\int_{0}^{\frac{1}{2}} \sin^{-1}(\sqrt{t})+\cos^{-1}(\sqrt{t}) \cdot{d}t$

$\left(\sin^{-1}(\sqrt{t})+\cos^{-1}(\sqrt{t})=\dfrac{\pi}{2}\right)$

$I=\displaystyle\int_{0}^{\frac{1}{2}} \dfrac{\pi}{2}\cdot{d}t$

$=\boxed{\dfrac{\pi}{4}}$