# Help with $\int\limits_{0}^{1}\int\limits_{2x}^{2}x^2\sin(y^4)\,dy\,dx$

I usually do my problems by myself and then check the solution with Wolfram Alpha, but in this situation, it's not helping me at all... I don't know if I got the wrong answer, or if wolfram is using some trig identity that I don't know of...

\begin{align} \int\limits_{0}^{1}\int\limits_{2x}^{2}x^2\sin(y^4)\,dy\,dx&=\int\limits_{0}^{2}\int\limits_{0}^{y/2}x^2\sin(y^4)\,dx\,dy\\ &=\frac{1}{3}\int\limits_{0}^{2}\left[x^3\sin(y^4)\right]_{x=0}^{x=y/2}\,dy\\ &=\frac{1}{24}\int\limits_{0}^{2}y^3\sin(y^4)\,dy\\ &=\frac{1}{96}\left[\sin(y^4)\right]_{y=0}^{y=2}\\ &=\frac{\sin(16)}{96} \end{align}

This is what I came up with, but wolfram is giving me the answer of $$\frac{\sin^2(8)}{48}$$

Needless to say, I'm not the best at remembering my trig identities. I didn't see anything on the wikipedia page of identities to give me any help either.

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The integral of sine is (minus) cosine. –  André Nicolas Aug 1 '13 at 1:43
@AndréNicolas Ah - that helps a bit, but my answer then would still be much different than wolfram's. But thanks to the explanations below, I now see the equality. –  agent154 Aug 1 '13 at 1:49
You would get $\frac{1}{96}(1-\cos 16)$. By the identity $\cos 2x=1-2\sin^2 x$, we have $1-\cos 16=2\sin^2 8$. –  André Nicolas Aug 1 '13 at 1:54

Your final antiderivative is incorrect. You should have

$$\int y^3 \sin{y^4} dy = -\cos{y^4}$$

$$\frac{1}{96} (-\cos(16) - (-\cos(0)) = \frac{1 - \cos{16}}{96}$$

As a sanity check, the numerical evaluation of this quantity agrees with Wolfram Alpha's evaluation of $\sin^2(8)/48$.

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You write $\frac{1}{96}\left[\sin(y^4)\right]_{y=0}^{y=2}$, but it should be $\frac1{96}[-\cos{y^4}]_0^4$.

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When you do the second integration, you should get: \begin{align} &=-\frac{1}{96}\left[\cos(y^4)\right]_{y=0}^{y=2}\\ &=-\frac{\cos(16)}{96}+\frac{1}{96} \end{align}

Now use the identity $\cos 2\theta=1-2\sin^{2}\theta$ and you should get the same answer as wolfram.

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