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March 27, 2015

March 27, 2015

Posted by **tom** on Sunday, February 20, 2011 at 7:56pm.

integral sin^4xdx

this the formula i used

integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2

using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral sin^2xdx= -1/2sinxcosx+1/2 integral 1 dx

can someone please show me how to combine the two integrals to get the final answer.

- calc -
**Helper**, Sunday, February 20, 2011 at 9:16pmMy answer, "|" = integral symbol

|sin^4 x dx =

-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 | sin x dx )

-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 (-cos x)) + C

-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x - 1/2 cos x) + C

-1/4 sin^3 x cos x - 3/8 sin x cos x

- 3/8 cos x + C

- calc -
**Helper**, Sunday, February 20, 2011 at 9:20pmHold on a few minutes, I might have a mistake.

- calc -
**Helper**, Sunday, February 20, 2011 at 9:52pmMistake fixed below

-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 | 1 dx )

-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 x) + C

-1/4 sin^3 x cos x - 3/8 sin x cos x

+ 3/8 x + C

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