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anonymous
 4 years ago
∫sin^5 t/cos^2 t dt
anonymous
 4 years ago
∫sin^5 t/cos^2 t dt

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First I split up both the sin^2x into 1cos^x So: I got ∫sinx(1 cos^2x)^2 / cos^2x Now, I use u substitution; So I let u = cosx du = sinxdx du = sinxdx Now I get: ∫(1 u^2)^2/ u^2 du expanding the top term ∫(1 2u^2 + 4u^2) / u^2 du now simplifying, ∫u^2 2 + 1/u^2 du Now, just integrate: and I got (1/3 u^3  2u  1/u) + c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(u^3/3  2u  1/u) + c\] \[((\cos^3x)/3) 2cosx 1/cosx) + c\]
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