Evaluate the integral $$\int x^7\cos(x^4)\,dx$$

I first tried integration by parts with $u = \cos(x^4)$, $du = -4x^3(\sin(x^4))$ $dv = x^7$ and $v = \frac 18 x^8$ which gave me

$$ \frac 18x^8\cos(x^4) - \int-4x^3\sin(x^4)\cdot\frac18x^8$$

I tried to solve the second integral using integration by parts, first plugging in $u = \frac 18x^8$ and $dv = -4x^3\sin(x^4)$. I discovered this doesn't work. I thought about other options for $u$ and $dv$ but it seemed like they would only yield more complicated equations, is there something I'm missing here? Can anyone guide me through the next step?


Start with the substitution $u=x^4$ to obtain $$ \int x^7\cos(x^4)\;dx=\int x^4\cos(x^4)\;x^3dx=\frac{1}{4}\int u\cos(u)\;du$$

Now integrate by parts, differentiating $u$ and integrating $\cos(u)$.


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