Hey, Veeky Forums. I tried evaluating this integral using trig sub, but it didn’t work. Can someone tell me why? (This is not a homework question, I just had a doubt)

# Hey, Veeky Forums

If you look closely there's a solution and explanation right below the question. Hope this helps!

I know, but I wanted to try it out with trig substitution and it didn’t work out. I don’t want to know the solution, I want to know why trig sub doesn’t work

I have this Phillips head screw and the right screwdriver right here, but I really want to use a wrench. Why won't it work?

square root x = tan^2 (theta)

You can get answers with more than one method. That's not a very good analogy.

it's not in the right form to use a trig sub. now go post this on /wsr/ or something. one calc II problem isn't really worth a thread

OP here. I just checked and trig sub does work. It gives you a different function, but the resulting graph is the same in both cases.

I got this result for your sub:

[math] \sqrt{x}= \tan^2 \theta[/math]

gives you the integral of

[math] \sin^3 \theta \cos \theta~d \theta[/math]

thats the problem, at this point you have to use a u-sub anyway, so why not just use it to begin with instead of a trig sub

just a demonstration of what Grandpa used to say:

there is more than one way to skin a cat

OP here. That's what I got. The final answer is sin^4 (theta)

It probably sounds weird, but its easier for me this way

The final answer is sin^4 (theta)

That's not the final answer though. How do you intend to bring this back into terms of [math]x[/math]?

You're saying [math] \int \frac{dx}{(1+\sqrt x)^3} = \sin^4 \theta[/math]. Your final answer shouldn't be in terms of [math] \theta [/math] when the question is in terms of [math]x[/math].

why the fuck would you simplify with trig?

just use x = u^2 to get it to be 2u/(1+u)^3, then add 0 = 2 - 2, getting (2u + 2)/(1+u)^3 - 2/(1+u)^3, which simplifies to 2/(1+u)^2 - 2/(1+u)^3. Then use the substitution u = w - 1 to get 2/w^2 - 2/w^3, from there, its absolutely basic

i have no idea why anybody would use trig on this problem