Question

Can someone check my work please. Evaluate the Integral

Answer 1

\[\int\limits 17 \sin^6x \cos^3x dx\]

Answer 2

This is what I got...\[17 \int\limits \sin ^6(x) \cos^2(x) \cos (x) dx= 17 \int\limits [(\sin^6(x)) (\sin^2(x)-1) (\cos (x))] dx\]=\[u= sin(x), du=-cos(x) dx, -du=cos(x) dx\] =\[17 \int\limits u^6 (u^2-1)(-du) = -17 \int\limits u^6(u^2-1) du\]=\[-17[\frac{ u^7 }{ 7 }(\frac{ u^3 }{ 3 }-x)]+C\]= \[17 [ \frac{ 1 }{ 7 } \sin (7x) +\frac{ 1 }{ 3 }\sin (3x)-x]+C\]=\[\frac{ 17 }{ 7 }\sin(7x)+\frac{ 17 }{ 3 }\sin(3x)-17x-C\]

Answer 3

i meant for the 2nd to last C to be - C not + C

Answer 4

https://www.wolframalpha.com/input/?i=integrate+17sinx^6cosx^3

Answer 5

wow. I will need a step by step on this one to see where i went wrong

Answer 6

keep in mind the sin(x)^n derives to n sin(x)^(n-1) cos(x)

so the idea is to get a cos(x) multiplied to a few sins
\[\sin^6x \cos^3x\]
\[(\sin^6x)(\cos^2x)\cos x\]
\[(\sin^6x)(1-\sin^2x)\cos x\]
\[(\sin^6x-\sin^8x)\cos x\]
integrate up to
\[\frac17\sin^7x-\frac19\sin^9x\]

Answer 7

@Jlockwo3 First thing I noticed was an error in the first line. \(\cos^2x=1-\sin^2x\), but \(\cos^2x\neq\sin^2x-1\).

Answer 8

Sorry for the extremely delayed response. I kept getting kicked off and gave up for the day! Anyways, this is what I ended up working out for this question....\[17 \int\limits u^6(1-u^2)du\] where u= cosx and du= -sinx dx\[= 17[\frac{ 1 }{ 7 }u^7-\frac{ 1 }{ 9 }u^9]\]\[=\frac{ 17 }{ 7 }\sin^7x-\frac{ 17 }{ 9 }\sin^9x+C\]

Answer 9

is this correct?

Answer 10

looks appropriate yes :)