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# I Need A Hint watch

1. INT: sqrt. tanx dx

Can anybody offer me a tip on what direction to proceed at the start from? (No complete solutions though please!)
2. tanx= sinx/cosx .
3. Integrate by substitution. Let u = sqrt(tanx).

Full solution below:
Spoiler:
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Let u = sqrt(tanx)
u2 = tanx

Differentiating,
2u du = [(secx)2] dx
2u du = (u2 + 1) dx
[2u / (u2 + 1)] du = dx

Integrate sqrt(tanx) dx,
= Integrate u[2u / (u2 + 1)] du
= Integrate [2(u^2) / (u2 + 1)] du
= Integrate {2 – [2 / (u2 + 1)]} du
= 2u – 2(tan-1u) + C
= 2[sqrt(tanx)] – 2{tan-1[sqrt(tanx)]} + C
4. Just use a subsitution such as u= tanx

(Original post by HanaanY)
tanx= sinx/cosx .
Thats quite useless in this case.
5. lol I was looking at my notes and I found this :
"What is the integral of tanx dx"
=INT (Sinx/Cosx)
=-ln(cosx) +c
=ln(cosx^-1) + c
Therefore, INT Tanx = ln(secx) +c

So that's why I thought it may be of some use. My bad - thats how bad I suck at maths
6. I think this is too difficult for me. I typed the expression into Mathematica Integrator and it is an incredible large answer (Not Knogle's)
7. (Original post by Knogle)
Integrate by substitution. Let u = sqrt(tanx).
Whatever you're smoking, I'd like a go.
(Original post by Malik)
Thats quite useless in this case.
No, it would just seem you have no idea what you're on about.

The hint is all you need. HanaanY, please don't apologise - you were much, much closer than any of the others. The integral is of the form -f'(x)/f(x), which is easy enough.
8. Yes, problem with substitution is you end up having to find dx/du of sqrt. tanx also.

I will try -f'(x)/f(x) form ideas.
9. (Original post by Wrangler)
Whatever you're smoking, I'd like a go.

No, it would just seem you have no idea what you're on about.
Excuse my ignorance but I still fail to see the use of tanx=sinx/cosx here?
10. Its the SQRT(tanx).

Am I being completely thick? I don't see how sqrt(sinx)/sqrt(cosx) is of form -f'(x)/f(x) ...
11. (Original post by e-unit)
Its the SQRT(tanx).

Am I being completely thick? I don't see how sqrt(sinx)/sqrt(cosx) is of form -f'(x)/f(x) ...
Neither do I.
12. -f'(x)/f(x)

Cos is the function, -sinx is its derivative

(for Int Tanx dx)
13. Here you go:

If you want a hint, don't read the entire thing.
14. Oh, jesus - looks like I owe a couple of you an apology. I didn't see the sqrt bit!!

15. (Original post by Knogle)
2u du = [(secx)2] dx
2u du = (u2 + 1) dx
That's not particularly correct. Think again about what u actually is.
16. Excellent dvs, thank you.

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