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,

$I=inttan^2xsec^4xdx$

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

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

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

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

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

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

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