Question

evaluate integral sin^3x cos^2xdx

Answer 1

change sin^3 to sin sin^2; sin^2 = 1-cos^2; and fit that to the cos^2 already there?

Answer 2

yup

Answer 3

\[\int sin\ sin^2\ cos^2\]
\[\int sin\ (1-cos^2)\ cos^2\]
\[\int sin\ (cos^2-cos^4)\]
\[\int (cos^2sin-cos^4sin)\]
perhaps?

Answer 4

yes but is there something that u= cosx and du= -sinx?

Answer 5

\[\int cos^2sin-\int cos^4sin\]

\[ -\frac{cos^3}{3}+ \frac{cos^5}{5}\]

Answer 6

let cox = z then dz = -sinx dx
now solve

Answer 7

sure, but only after you split it into workable parts

Answer 8

ok thanks!! big help : )

Answer 9

i think the wolfram disagrees with my assessment tho

Answer 10

so i could have made a mistake somewhere

Answer 11

\(\int\limits_{}^{}\sin^3{x} \cos^2{x}dx=\int\limits_{}^{}\sin{x}(\cos^2x-\cos^4{x})dx\). Use a substitution \(u=\cos{x} \implies du=-\sin{x}dx\). So the integral becomes \(\int\limits_{}^{}(u^4-u^2)du={u^5 \over 5}-{u^3 \over 3}+c\). Now, just substitute back \(u=\cos{x}\).

Answer 12

no its good, they just take it another step farther

Answer 13

(1-cos^2 x)cos^2 x (-dz)
=> -(1-z^2)(z^2)dz
=> (-z^2 + z^4)dz
=> -z^3/3 + z^5/5
=> - cos^3/3 + cos^5/5 +c

Answer 14

I think I am too slow! -.-

Answer 15

the wolf give ...which i do not see as simplier by any means

1/30 cos^3(x) (3 cos(2 x)-7)+c

Answer 16

its not that there is something to do with a u-sub; there is nothing magical about the usub other than it tends to clean up the clutter