How do I find #inttan^4x dx#?

1 Answer
Gió
Mar 7, 2015

Well, this one is quite difficult;

#inttan^4(x)dx=int tan^2(x)tan^2(x)dx=#
#=intsin^2(x)/cos^2(x)tan^2(x)dx=#
#=int(1-cos^2(x))/cos^2(x)tan^2(x)dx=#
#=int[1/cos^2(x)-1]tan^2(x)dx=#
#=int[tan^2(x)/cos^2(x)-tan^2(x)]dx=#
But #d[tan(x)]=1/cos^2(x)dx#

#=inttan^2(x)d[tan(x)]-inttan^2(x)dx=#
#=tan^3(x)/3-inttan^2(x)dx=#
#=tan^3(x)/3-intsin^2(x)/cos^2(x)dx=#
#=tan^3(x)/3-int[(1-cos^2(x))/cos^2(x)]dx=#
#=tan^3(x)/3-{int1/cos^2(x)dx-intdx}=#
#=tan^3(x)/3-tan(x)-x+c#

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