Integral $\int x^7\cos x^4 dx$
Solution 1
make a substitution $u = x^4, du = 4x^3 dx.$ the $$\int x^7 \cos x^4 \, dx = \frac14\int u\cos u\, du = \frac14 \int u \, d(\sin u) = \frac14 \left( u\sin u -\int \sin u \,du\right) =\frac14 \left( u\sin u + \cos u\right) + C = \frac14 \left( x^4\sin x^4 + \cos x^4\right) +C $$
Solution 2
Hint
$$I=\int x^7 \cos(x^4)\,dx=\frac 14\int (4x^3)\, x^4 \cos(x^4)\,dx$$ So, let $x^4=u$ and then $$I=\frac 14\int u \cos(u) \, du$$ I am sure that you can take from here.
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King Squirrel
Updated on July 31, 2022Comments
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King Squirrel over 1 year
$\displaystyle \int x^7\cos x^4 dx$
I tried first by letting $x^4 = u$ and then using integration by parts by assigning f(x) to $u^\frac74$ and cos(u) to g'(x) and I end up getting after applying parts twice, the same integral on the RHS as what we are looking for. So I bring it in on the LHS and add it over and get $\displaystyle \cos x^4 \bigg (\frac{\displaystyle 4 \displaystyle u^\frac{11}{4}}{11} \bigg)$
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Patrick Da Silva over 8 yearsYou should switch back your $u$ to $x^4$ at the end, otherwise +1.
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abel over 8 yearsgood to see you claude.
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Claude Leibovici over 8 yearsThanks, abel. It is also my pleasure to meet you quite often here. Cheers :-)