Question 356868
Let {{{u=3x}}}, {{{du=3dx}}}
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( sin^2(u)*cos^5(u), du )}}}
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{{{(sin(u))^2+(cos(u))^2=1}}}
{{{(cos(u))^2=1-(sin(u))^2}}}
{{{(cos(u))^4=1-(sin(u))^4}}}
Substitute,
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( sin^2(u)*cos^4(u)*cos(u), du )}}}
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( sin^2(u)*(1-(sin(u))^4)*cos(u), du )}}}
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Let {{{v=sin(u)}}}, {{{dv=cos(u)du}}}
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( v^2*(1-v^2)^2, dv)}}}
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( v^2*(1-2v^2+v^4), dv)}}}
{{{int ( sin^2(3x)*cos^5(3x), dx )=3*int ( (v^2-2v^4+v^6), dv)}}}
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Now the problem is reduced to a simple polynomial integration.
Integrate and then substitute back for v and then finally u.
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Additionally, this is not an algebra problem.