[tex]\int sin^3(4x)cos^{10}(4x) dx[/tex](adsbygoogle = window.adsbygoogle || []).push({});

ok i know that i need to borrow the a sin, cause the sin has an odd power

[tex]\int sin^2(4x)cos^{10}(4x) sin(4x)dx[/tex]

ok used one of the trig ids. on [tex]sin^2(4x)[/tex]

= [tex]\int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx[/tex]

=[tex]\int cos^{10}(4x)*sin(4x)dx - \int cos^{12}(4x)cos^2(4x)sin(4x)dx[/tex]

ok so i start to integrate the left side first...

[tex] \int sin(4x)cos^{10}(4x)dx [/tex]

[tex]u=cos(4x)[/tex]

[tex]du=-4sin(4x)dx[/tex]

[tex] dx=-1/4 [/tex] <-- needs to be balanced right? man im so confused on balancing, what am i suppose to balance anyway? i usally guess it lol, which is not good. please help me with this also

so...

[tex] \int -1/4 du * u^{10} [/tex]

[tex] -1/4 \int u^{10} * du [/tex]

now to find the anti-derv.

[tex] -1/4(u^{11}/11) [/tex]

sub back in for u

[tex] -1/4(cos(4x)^{11}/11) [/tex]

so this is what the problem looks right now...

[tex] -1/4(cos(4x)^{11}/11) - \int cos^{12}(4x)cos^2(4x)sin(4x)dx[/tex]

ok how would i do the right side? it's so huge, so i use trig ids. agian? but am i doing it right so far?

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# Homework Help: Math help

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