Integrate #int_0^(pi/2)sin^7x*cos^5xdx# ?

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Answer 1 · 2018-06-28T13:08:52

The answer is #=1/120#

Apply

#cos^2x=1-sin^2x#

Calculate the indefinite integral first

Therefore,

The integral is

#I=intsin^7xcos^5xdx=intcosxsin^7x(1-sin^2x)^2dx#

Let #u=sinx#, #=>#, #du=cosxdx#

Therefore,

#I=intu^7(1-u^2)^2du#

#=int(u^11-2u^9+u^7)du#

#=u^12/12-2u^10/10+u^8/8#

#=sin^12x/12-sin^10x/5+sin^8/8+C#

The definite integral is

#int_0^(pi/2)sin^7xcos^5xdx= [sin^12x/12-sin^10x/5+sin^8/8]_0^(pi/2)#

#=(1/12-1/5+1/8)-(0)#

#=1/120#