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I came across this integral in a math competition and couldn't solve it $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x})\, dx$$

I tried a $u$-substitution, with $u=\sin x$ and ended up with the integral $$\int_0^\frac{\pi}{4} \cos^{-1}(\text{u}) \cdot \frac{1}{\sqrt{1-u^2}}\,du$$ which is not much simpler and I cannot figure out how to solve this. Any hints/solutions for this problems?

I also tried drawing a triangle for the problem but it didn't really help with the solution.

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5 Answers 5

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You can use the trigonometric formula : $$\forall x \in\mathbb{R},\cos^{-1}(x)+\sin^{-1}(x)=\frac{\pi}{2}$$

Thus $\forall x\in\left[0,\frac{\pi}{4}\right]$ you have $\sin^{-1}(\sin(x))=x$ so : $$\cos^{-1}(\sin(x))+\sin^{-1}(\sin(x))=\frac{\pi}{2}\Rightarrow \cos^{-1}({\sin x})=\frac{\pi}{2}-x$$

Finally : $$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) dx=\int_0^\frac{\pi}{4} \frac{\pi}{2}-xdx$$ Can you finish ?

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  • $\begingroup$ What is $\sin^{-1}(\sin x)=?$ $\endgroup$ Commented Apr 11, 2017 at 15:48
  • $\begingroup$ it is $x$ since $x$ is in $[0,\pi/4]$ $\endgroup$
    – Bérénice
    Commented Apr 11, 2017 at 15:54
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HINT:

If $\cos^{-1}(\sin x)=y,0\le y\le\pi\ \ \ \ (1)$

and $\cos y=\sin x=\cos\left(\dfrac\pi2-x\right)$

$y=2m\pi\pm\left(\dfrac\pi2-x\right)$ where $m$ is an integer such that $(1)$ is satisfied

For $0\le2m\pi+\dfrac\pi2-x\le\pi\iff 2m\pi-\dfrac\pi2\le x\le2m\pi+\dfrac\pi2$

Here $m=0$

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Here's a barely different route to take, continuing from the direction you've taken.

First, note that substituting $u=\sin x$ would actually give

$$\int_{x=0}^{x=\pi/4}\cos^{-1}(\sin x)\,\mathrm dx=\int_{\color{red}{u=0}}^{\color{red}{u=1/\sqrt2}}\frac{\cos^{-1}u}{\sqrt{1-u^2}}\,\mathrm du$$

Now, recall that $\dfrac{\mathrm d}{\mathrm du}\cos^{-1}u=-\dfrac1{\sqrt{1-u^2}}$, which means you can make another intermediate substitution of, say, $v=\cos^{-1}u$. Then you have

$$-\int_{v=\cos^{-1}0=\pi/2}^{v=\cos^{-1}(1/\sqrt2)=\pi/4}v\,\mathrm dv=\int_{\pi/4}^{\pi/2}v\,\mathrm dv=\dfrac{3\pi^2}{32}$$

which agrees with the other answers above.

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$$\int \cos^{-1}(\sin x) dx$$ $$= \int 1 \cdot \cos^{-1}(\sin x) dx = x \cdot \cos^{-1}(\sin x) - \int x \cdot -\frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ (integration by parts)

$$ = x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C $$ $$ = x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{\cos^2 x}} dx + C $$

$$= x \cdot \cos^{-1}(\sin x) \pm \int x \cdot \frac{\cos x}{\cos x} dx + C$$ $$ = x \cdot \cos^{-1}(\sin x) \pm \frac{x^2}{2} + C$$

Now do it for the definite integral.

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1)

$(u^2)^{\prime}=2\times u\times u^{\prime}$

therefore $\dfrac{1}{2}u^2$ is an antiderivative of $u\times u^{\prime}$

2) derivative of $\text{cos}^{-1}(x)=-\dfrac{1}{\sqrt{1-x^2}}$

$\displaystyle J= \int_0^{\tfrac{\pi}{4}} \text{cos}^{-1}(\sin x)dx$

Perform the change of variable $y=\sin x$,

$\displaystyle J=\int_0^{\tfrac{\sqrt{2}}{2}} \dfrac{\text{cos}^{-1}(x)}{\sqrt{1-x^2}}dx$

Using $(1)$,

$\begin{align}J&=-\dfrac{1}{2}\Big[\text{cos}^{-1} (x)^2\Big]_0^{\tfrac{\sqrt{2}}{2}}\\ &=-\dfrac{1}{2}\left[\dfrac{\pi^2}{16}-\dfrac{\pi^2}{4}\right]\\ &=\boxed{\dfrac{3}{32}\pi^2} \end{align}$

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