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What is the integral of #int sin^5x*cos^7x #?

1 Answer

Aug 25, 2016

#-(1/8cos^8 x-2/10 cos^10 x + 1/12 cos^12 x) + C#

Explanation:

# sin^5x*cos^7x = sin^4x cos^7x sinx = (1-cos^2 x)^2cos^7x sin x#
#=(1-2cos^2x+cos^4 x)cos^7 x sin x #
#=(cos^7x-2 cos^9 x+cos^11 x)sinx# so

#int sin^5x*cos^7x dx = int (cos^7x-2 cos^9 x+cos^11 x)sinx dx#
#=-(1/8cos^8 x-2/10 cos^10 x + 1/12 cos^12 x) + C#

or simplifying

#-1/480(27 - 28 cos(2 x) + 5 cos(4 x)) cos^8x#

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