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For the following integral problem, we must show the sum of the 2 integrals is equal to $\pi/4$.

$$\int_{0}^{\sin^2x}\sin^{-1}(\sqrt{t})\ dt \; +\int_{0}^{\cos^2x}\cos^{-1}(\sqrt{t})\ dt= \frac{\pi}{4} $$

We must use the Fundamental Theorem of Calculus to resolve this problem, because calculating this integral takes far too long to do!

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    By "$sinx^2$", do you actually mean $\sin x^2=\sin(x^2)$, or do you mean $\sin^2x=(\sin x)^2$? – 2017-02-13
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    its $$sin^2x=(\sin x)^2$$ – 2017-02-13

3 Answers 3

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Sketch of proof:

  1. Derive the function $$g(x)=\int_0^{\sin^2 x}\arcsin\sqrt t\,dt+\int_0^{\cos^2 x}\arccos\sqrt t\,dt$$ to obtain that $g'(x)= 0$ for all $x\ne \frac k2\pi$ and, hence, that $g$ is constant.

  2. Look for a (easy) real number $\xi>0$ such that $\sin^2\xi=\cos^2\xi$.

  3. Use the identity $\arccos x=\frac\pi2-\arcsin x$ to obtain $$g(\xi)=\int_0^{\sin^2 \xi}\arcsin\sqrt t\,dt+\int_0^{\cos^2 \xi}\arccos\sqrt t\,dt=\\=\int_0^{\sin^2\xi}\arcsin\sqrt t+\arccos\sqrt t\,dt=\frac\pi2\sin^2\xi$$

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Introduce a change of variable for the first integral \begin{align*} u & = \arcsin\sqrt{t} \\ t & = (\sin u)^2 \\ dt & = 2(\cos u)(\sin u) du = \sin(2u) du \end{align*} and for the second integral \begin{align*} u & = \arccos\sqrt{t} \\ t & = (\cos u)^2 \\ dt & = - 2(\cos u)(\sin u) du = - \sin(2u) du \end{align*} Then, for $x \in [0,\pi/2]$ we have \begin{align*} g(x) & = \int_0^x u \sin(2u) du + \int^{\pi/2}_x u \sin(2 u) du \\ & = \int_0^{\pi/2} u \sin(2u) du \\ & = \left[\frac{1}{4} \sin(2u) - \frac{1}{2} u\cos(2u) \right]_0^{\pi/2}\\ &= \frac{\pi}{4}. \end{align*} Since from the definition of $g$ it follows that $g(\pi/2+x) = g(\pi/2-x)$ for all $x \in \mathbb R$, we have $g(x) = \pi/4$ for all $x \in [0,\pi]$. As $g$ is clearly periodic with the period $\pi$ we have $g(x) = \pi/4$ for all $x \in\mathbb R$.

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For all $x \in [0,\pi/2]$ we have the identity $(\cos x)^2 = 1-(\sin x)^2$. Since both functions in this identity are bijections between $[0,\pi/2]$ and $[0,1]$ for their inverses we have $\arccos\sqrt{y} = \arcsin\sqrt{1-y}$ for all $y \in [0,1]$. Using this identity in the definition of $g$ we have \begin{align*} g(x) & = \int_0^{(\sin x)^2} \arcsin\sqrt{t}\, dt + \int_0^{(\cos x)^2} \arcsin\sqrt{1-t}\, dt \\ & = \int_0^{(\sin x)^2} \arcsin\sqrt{t}\, dt + \int_{1-(\cos x)^2}^1 \arcsin\sqrt{s} \, ds \\ & = \int_0^{1} \arcsin\sqrt{t} \,dt \\ & = \int_0^{\pi/2} u \sin(2u) dt \\ & = \pi/4. \end{align*}