Question

NEED HELP.... Calculate the integration of sin^3(X)cos^6(X)dx....

Answer 1

\[\int \sin^3(x)\; \cos^6(x) \;dx\]
Looks like chain rule may be involved. Have you tried u-substitution?

Answer 2

i did the u subsitution...but im stuck on it .. no idea where i went wrong

Answer 3

maybe we need to reduce the power on cos a little. Some trig identity???

Answer 4

wouldnt you reduce the sin since its an odd exponent..

Answer 5

then use cos^2x+sin^2x=1....

Answer 6

yah,
\[\sin^3(x) = \sin(x)\sin^2(x) = \sin(x) (1-cos^2(x))\]
so at least one part of the integral can be dealt with...

Answer 7

i got = (1-u^2)(u^6)(-du).....

Answer 8

that was for the subsitution before integrating... \

Answer 9

u = cosx....du=-sinx

Answer 10

ok...

Answer 11

did some distributing before i integrateded... and got u^12-u^6 du...

Answer 12

??? I got
\[\int (u^6 - u^8) du\]

Answer 13

ohhhhh you add exponents ... thats right... forgot

Answer 14

:)

Answer 15

that was my flaw.... thank you

Answer 16

cool