# Thread: a couple of trigonometric integrals

1. ## a couple of trigonometric integrals

$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$

My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.

2. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
Here's how you do the first one:

Spoiler:
Let

$\displaystyle \displaystyle I=\int_0^\pi x\sin\left(\cos^2(x)\right)\cos^2(\sin(x))\text{ }dx$

Then, letting $\displaystyle z=\pi-x$ we find that

\displaystyle \displaystyle \begin{aligned}I &=\int_0^\pi (\pi-z)\sin\left(\cos^2(z)\right)\cos\left(\sin^2(z)\ri ght)\text{ }dz\\ &=\pi\int_0^\pi\sin\left(\cos^2(z)\right)\cos\left (\sin^2(z)\right)\text{ }dz-I\end{aligned}

Thus,

$\displaystyle \displaystyle \frac{2}{\pi}I=\int_0^\pi \sin\left(\cos^2(z)\right)\sin\left(\cos^2(z)\righ t)\text{ }dz$

Recall though that

$\displaystyle \displaystyle \sin(\phi)\cos(\theta)=\frac{\sin(\phi+\theta)+\si n(\phi-\theta)}{2}$

And thus

\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi} I &= \int_0^\pi \sin\left(\cos^2(z)\right)\cos^2\left(\sin(z)\righ t)\text{ }dz\\ &= \frac{1}{2}\int_0^\pi\left(\sin\left(\cos^2(z)+\si n^2(z)\right)+\sin\left(\cos^2(z)-\sin^2(z)\right)\right)\text{ }dz\\ &=\frac{1}{2}\pi\sin(1)+\frac{1}{2}\int_0^\pi \sin(\cos(2z))\text{ }dz\end{aligned}

Note though that

$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^\pi\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))\text { }dz$

Note though that by letting $\displaystyle \displaystyle u=z-\frac{3\pi}{4}$ one gets that

$\displaystyle \displaystyle \int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))=\int_0^{ \frac{\pi}{4}}\sin(\cos(2u))\text{ }du$

And thus

$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{\pi}{4}}^{\frac{3\pi}}{4}\sin(\cos (2z))\text{ }dz$

But, letting $\displaystyle w=z-\frac{\pi}{2}$ in this second integral one arrives at

$\displaystyle \displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z) )=-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$

and since the integrand is even this is equal to

$\displaystyle \displaystyle -2\int_0^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$

Therefore,

\displaystyle \displaystyle \begin{aligned}\int_0^\pi\sin(\cos(2z))\text{ }dz &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))+\int_{\frac{ \pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz-2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &=0\end{aligned}

Thus, finally we arrive at

\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi}I &=\frac{\pi}{2}\sin(1)+\frac{1}{2}\int_0^\pi\sin(\ cos(2z))\text{ }dz\\ &=\frac{\pi}{2}\sin(1)\end{aligned}

and thus $\displaystyle \displaystyle I=\frac{\pi^2 \sin(1)}{4}$.

3. The ODE you had was solvable.

4. Originally Posted by Random Variable
The ODE you had was solvable.
See my new method. It's more elementary.

5. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$

My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
The second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet?

6. to prove that $\displaystyle \displaystyle\int_0^\pi\sin(\cos 2x)\,dx=0,$ split it up into two integrals with $\displaystyle 0\le x\le\dfrac\pi2$ and $\displaystyle \dfrac\pi2\le x\le\pi$ then apply substitution $\displaystyle t=x-\dfrac\pi2$ on the second piece and you'll see is equal to minus the first piece, hence, the result.

7. Here's my solution to the second

Spoiler:

First use the double angle formula so

$\displaystyle \displaystyle \int_0^{\pi} x\left( \dfrac{1-\cos \left(2 \sin x\right)}{2} + \dfrac{1 + \cos \left(2 \cos x\right)}{2} \right)\, dx$

$\displaystyle \displaystyle = \int_0^{\pi} x\,dx + \frac{1}{2}\int_0^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$

We will show the second integral is zero by considering it in two parts. (**)

First, that $\displaystyle \displaystyle \int_0^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Let $\displaystyle x = \pi -u$ in $\displaystyle \displaystyle \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Next that $\displaystyle \displaystyle \int_0^{\pi/4} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = -\int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Let $\displaystyle x = \pi/2-u$ in$\displaystyle \displaystyle \int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$

Thus, we are left with $\displaystyle \displaystyle \int_0^{\pi} x\,dx = \dfrac{\pi^2}{2}$

8. here's mine to the second:

put $\displaystyle t=\pi-x$ on the integral to see that is equal to $\displaystyle \displaystyle\frac\pi2\int_0^\pi\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx.$

i claim the latter integral equals $\displaystyle \pi,$ hence the original integral equals $\displaystyle \dfrac{\pi^2}2,$ so in order to see that write the integral as

$\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx},$

and put $\displaystyle t=x-\dfrac\pi2$ on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.

another solution for the first one:

we consider $\displaystyle \displaystyle\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ (1) then write it as $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}$ and put $\displaystyle t=x-\dfrac\pi2$ on the second one then we get $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)\,dx},$ so when adding those we note that the integrand is actually $\displaystyle \sin(\sin^2x+\cos^2x)$ so (1) equals $\displaystyle \dfrac\pi2\sin1.$

on the original integral was easy to show that it's equal to $\displaystyle \displaystyle\frac\pi2\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ thus by (1) we conclude.