What is the Integral of #tan^2 x sec^4 x dx#?

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Answer 1 · 2018-05-27T14:30:33

#I=tan^3x/3+tan^5x/5+c#

We know that,

#(1)color(red)(int[f(x)]^nd/(dx)(f(x))dx=[f(x)]^(n+1)/(n+1)+c, where,(n!=-1,f(x)>0 and f'(x)!=0)#

We have,

#I=inttan^2xsec^4xdx#

#=int tan^2x(sec^2x)sec^2xdx#

#=inttan^2x(1+tan^2x)sec^2xdx#

#=int tan^2xsec^2xdx+inttan^4xsec^2xdx#

#=int(tanx)^2d/(dx)(tanx)dx+int(tanx)^4d/(dx)(tanx)dx#

#=(tanx)^3/3+(tanx)^5/5+c...to Apply(1)#

#=tan^3x/3+tan^5x/5+c#