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integratation of power of sines and cosines: ∫ sin^4 xcos^2xdx

Calculus1
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this is calculus
\[-\frac{ \cos ^{5}x }{ 5 }+\frac{ \cos ^{7}x }{ 7 }+c\]
how come? what case did you used? is it case 2?

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Other answers:

bak reyiz, cos^2 x=1-sin^2 x. biliyon mu bunu??
what? im sorry i cant understant your language, could you speak in english please?
can someone please help me? :(
I think it's not like what marsss did there... \[∫ sin^4 xcos^2xdx\]\[=∫ sin^4 x(1-sin^2x)dx\]\[=∫ sin^4x-sin^6xdx\]\[=∫ (sin^2x)^2dx-∫(sin^2x)^3dx\]\[=\frac{1}{4}∫ (1-cos2x)^2dx-\frac{1}{8}∫(1-cos2x)^3dx\]\[=\frac{1}{4}∫ (1-2cos2x+cos^22x)dx-\frac{1}{8}∫(1-3cos2x+3cos^22x-cos^32x)dx\] Reduce the power of cosine into 1 using doucle angle formula, except for cos^3(2x). Good luck :|

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