How do you integrate $(tanx)^5*(secx)^4dx$ ?

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Answer 1 · 2016-11-26T22:53:09

$tan^8x/8+tan^6x/6+C$

Writing this as:

$inttan^5xsec^4xdx$

Rewriting to leave all $tanx$ functions with one cluster of $sec^2x$ , which is the derivative of $tanx$ :

$=inttan^5xsec^2xsec^2xdx$

$=inttan^5x(tan^2x+1)sec^2xdx$

Letting $u=tanx$ so $du=sec^2xdx$ :

$=intu^5(u^2+1)du$

$=intu^7du+intu^5du$

$=u^8/8+u^6/6+C$

$=tan^8x/8+tan^6x/6+C$

How do you integrate (tanx)^5*(secx)^4dx(tanx)^5*(secx)^4dx?