Hello. Could you help me integrate this using trig sub? Thank you. I was stuck on a few steps. What does a = ? x = ? dx = ? `intx^3sqrt(x^2-25).`

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nick-teal | High School Teacher | (Level 3) Adjunct Educator

Posted on

Ok, this is gonna take a while.

The trig substitution we will make depends on how the stuff underneath the radical is arranged. Check out the table in the link I am posting below.

For the form `sqrt(x^2 - a^2)`

We use the substitution `x = asec(theta)`

In this case a = 5, so our substitution is `x = 5sec(theta)`

Then we derive to find `dx` and `d theta`

`dx = 5sec(theta)tan(theta) d theta`

Plug this into our integral at we get 

`int (5sec(theta))^3 * sqrt((5sec(theta))^2 - 5^2) * 5 sec(theta)tan(theta) d theta`

We can factor out a 5^2 from each term under the radical to turn that into

`sqrt(5^2(sec(theta)^2 - 1))`

And using our trusty trig identities this turns into

`sqrt(5^2 * tan^2(theta))`

We will rewrite out equation without the radical now.

`int (5sec(theta))^3 * 5tan(theta) * 5sec(theta)tan(theta) d theta`

We combine like terms to get

`int 3125 sec^4(theta) tan^2(theta) d theta`

At this point we will want to solve our new integral using u substitution.

We will do this by turning sec^4 into sec^2*(tan^2 + 1) 

`3125 int sec^2(theta) (tan^2(theta) + 1) tan^2(theta) d theta`

Now we let `u = tan(theta)`

so `du = sec^2 (theta) d theta`

Plug this all in, and we get

`3125 int (u^2 + 1)u^2 du`

integrate this to get 

`3125(u^5/5 + u^3/3) + c` .

Now all we need to do is sub back in the two substitutions we made and we are finished!

`3125((tan^5(theta))/5 + (tan^3(theta))/3) + c`

Now sub in `x = 5sec(theta)`

But we do not have secants, we have tangents in our equation!

We will have to use triangle trig to figure out what tangent equals.

`x = 5sec(theta)` 

`sec(theta) = x/5` so `cos(theta) = 5/x`  

Adjacent = 5, hypotenuse = x, which implies that opposite = `sqrt(x^2 - 25)`

This means that `tan(theta) = sqrt(x^2 - 25) / 5`

Now we can finally plug that in for our tangents, and we get

`3125((sqrt(x^2-25)/5)^5/5 + (sqrt(x^2 - 25)/5)^3/3) + c`

And you can have fun simplifying that on your own.

Hope this helps!!!

Sources:
embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

`intx^3sqrt(x^2-25)dx` using a trigonometric substitution.

Let `x=5sectheta,dx=5secthetatantheta d theta`

`sqrt(x^2-25)=5tantheta `

`intx^3sqrt(x^2-25)dx=int(125sec^3theta)(5tantheta)(5secthetatantheta d theta)`

`=3125inttan^2theta(1+tan^2theta)sec^2theta d theta`

`=3125((tan^3 theta)/3+(tan^5 theta)/5)+C`

`=3125/15tan^3theta[5+3tan^2theta]+C`

`=625/3((x^2-25)^(3/2))/125[5+3(x^2-25)/25]+C`

`=1/15(x^2-25)^(3/2)[125+3(x^2-25)]+C`

`=1/15(x^2-25)^(3/2)(3x^2+50)+C`

Sources:

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moocow554 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

Hello. Thank you for your explanation. Is there anyway you could be more specific in your steps? Maybe some background info? Thanks

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