Question

How do you integrate `int t^5/sqrt(t^2+2)` by trigonometric substitution?

Answer 1

`(sqrt(t^2+2)(3t^4-8t^2+32))/15+C`

Explanation:

`I=intt^5/sqrt(t^2+2)dt`

Let `t=sqrt2tantheta`. This implies that `dt=sqrt2sec^2thetad theta`. Plugging these in yields

`I=int(sqrt2tantheta)^5/sqrt(2tan^2theta+2)(sqrt2sec^2thetad theta)`

`I=(sqrt2)^5int(tan^5thetasec^2theta)/sqrt(tan^2theta+1)d theta`

Since `tan^2theta+1=sec^2theta`:

`I=4sqrt2inttan^5thetasecthetad theta`

`I=4sqrt2inttan^4theta(tanthetasectheta)d theta`

`I=4sqrt2int(sec^2theta-1)^2(secthetatantheta)d theta`

Letting `u=sectheta` so `du=secthetatanthetad theta`

`I=4sqrt2int(u^2-1)^2du=4sqrt2int(u^4-2u^2+1)du`

`I=4sqrt2(u^5/5-2/3u^3+u)`

`I=4sqrt2sectheta(sec^4theta/5-2/3sec^2theta+1)`

Our original substitution was `tantheta=t/sqrt2`. Thus `sectheta=sqrt(tan^2theta+1)=sqrt(t^2/2+1)=1/sqrt2sqrt(t^2+2)`.

`I=4sqrt2(1/sqrt2sqrt(t^2+2))(1/5(sqrt((t^2+2)/2))^4-2/3(sqrt((t^2+2)/2))^2+1)`

`I=4sqrt(t^2+2)(1/5((t^2+2)/2)^2-2/3((t^2+2)/2)+1)`

`I=4sqrt(t^2+2)((t^4+4t^2+4)/20-(t^2+2)/3+1)`

`I=4sqrt(t^2+2)((3t^4+12t^2+12)/60-(20t^2+40)/60+60/60)`

`I=(sqrt(t^2+2)(3t^4-8t^2+32))/15+C`

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However, note that a trig substitution was not actually necessary.

`I=intt^5/sqrt(t^2+2)dt`

Let `u=t^2+2` so `du=2tdt`. This also implies that `t^4=(u-2)^2`.

`I=1/2intt^4/sqrt(t^2+2)(2tdt)=1/2int(u-2)^2/sqrtudu`

Expanding the numerator and dividing each term, then integrating term by term will give the same result with less hassle.