`int sec^5x tan^3x dx` Find the indefinite integral

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`intsec^5(x)tan^3(x)dx`

Let's rewrite the integral as:

`intsec^5(x)tan^3(x)dx=intsec^5(x)tan^2(x)tan(x)dx`

Now using the trigonometric identity:`tan^2(x)=sec^2(x)-1`

`=intsec^5(x)(sec^2(x)-1)tan(x)dx`

`=intsec^4(x)(sec^2(x)-1)sec(x)tan(x)dx`

Now apply the integral substitution:`u=sec(x)`

`du=sec(x)tan(x)dx`

`=intu^4(u^2-1)du`

`=int(u^6-u^4)du`

apply the sum rule,

`=intu^6du-intu^4du`

`=(u^(6+1)/(6+1))-(u^(4+1)/(4+1))`

`=u^7/7-u^5/5`

substitute back `u=sec(x)` and add a constant C to the solution,

`=(sec^7(x))/7-(sec^5(x))/5+C`

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`intsec^5(x)tan^3(x)dx`

Let's rewrite the integral as:

`intsec^5(x)tan^3(x)dx=intsec^5(x)tan^2(x)tan(x)dx`

Now using the trigonometric identity:`tan^2(x)=sec^2(x)-1`

`=intsec^5(x)(sec^2(x)-1)tan(x)dx`

`=intsec^4(x)(sec^2(x)-1)sec(x)tan(x)dx`

Now apply the integral substitution:`u=sec(x)`

`du=sec(x)tan(x)dx`

`=intu^4(u^2-1)du`

`=int(u^6-u^4)du`

apply the sum rule,

`=intu^6du-intu^4du`

`=(u^(6+1)/(6+1))-(u^(4+1)/(4+1))`

`=u^7/7-u^5/5`

substitute back `u=sec(x)` and add a constant C to the solution,

`=(sec^7(x))/7-(sec^5(x))/5+C`

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