How do I evaluate the integral #intsec^3(x) tan(x) dx#?

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Answer 1 · 2015-02-18T14:17:41

I would start by writing your integrand as:

#int1/cos^3(x)*sin(x)/cos(x)dx=intsin(x)/cos^4(x)dx=#

Now: #d[cos(x)]=-sin(x)dx#

I can write the integral in a new equivalent form:

#-int(d[cos(x)])/cos^4(x)=# now you can use #cos(x)# as if it were a simple #x# during your integration, giving:

#-int(d[cos(x)])/cos^4(x)=-intcos^(-4)(x)d[cos(x)]=#
(as for #-intx^-4dx#)

#=cos^(-3)/3+c=1/(3cos^3(x))+c#