How do you find the definite integral of #t^7cos(t^4) # from #[pi^(1/4), (pi/2)^(1/4)]#?

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Answer 1 · 2016-09-07T13:49:24

#1/8(pi+2)#.

Let #I=int_(pi^(1/4))^((pi/2)^(1/4)) t^7cos(t^4)dt#.

We subst. #t^4=x rArr 4t^3dt=dx#

Also, #t=(pi)^(1/4) rArr x=pi, and, t=(pi/2)^(1/4) rArr x=pi/2#. Therefore,

#I=1/4int_(pi) ^ (pi/2)xcosxdx#

We first find the Indefinite Integral #intxcosxdx# using IBP :

#intxcosxdx=x*intcosxdx-int[d/dx(x)*intcosxdx]dx#

#=xsinx-intsinxdx#

#=xsinx-(-cosx)#

#xsinx+cosx+C#

Therefore, #I=1/4[xsinx+cosx]_(pi)^(pi/2)#

#=1/4[{pi/2*sin(pi/2)+cos(pi/2)}-{pisin(pi)+cos(pi)}]#

#=1/4(pi/2+1)#

#=1/8(pi+2)#.