The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Fp2 Integration



i hope that's come up right ^
basically, how on earth do you do that integration?? i can't get my head round it. i tried an alternative way, of taking out a (x^2)^1/2 and got a similarish integral of 1/2(1+(9/4x^2))^1/2 (i had a different multiple of what they were integrating, but if you multiplied mine by 2/2 then you got the same)
but it didn't work. please help me!! (:
Reply 1
Avatar for Laurenje
Laurenje
They've let u = arcsinh(2x/3) and dv/dx = 1 and then used integration by parts. Does this help?
Reply 2
Avatar for Josephiiine
Josephiiine
OP
i got that bit, it's just how they got 2x/(9+4x^2)^1/2 into... -1/2(9+4x^2)^0.5 (the bit on the mark scheme which is under A1 ft)
Josephiiine
...


It's done on sight as it is presence of the derivative... consider that

ddx[9+4x2]=129+4x2×ddx[4x2]=8x29+4x2 \dfrac{d}{dx}[\sqrt{9+4x^2}] = \dfrac{1}{2\sqrt{9+4x^2}} \times \dfrac{d}{dx}[4x^2] = \dfrac{8x}{2\sqrt{9+4x^2}}

Otherwise if you don't like that method, make trig/hyperbolic substitutions in those situations - I would suggest trying x=32sinh⁡t x = \frac{3}{2} \sinh t in this case.
spread_logic_not_hate
It's done on sight as it is presence of the derivative... consider that

ddx[9+4x2]=129+4x2×ddx[4x2]=8x29+4x2 \dfrac{d}{dx}[\sqrt{9+4x^2}] = \dfrac{1}{2\sqrt{9+4x^2}} \times \dfrac{d}{dx}[4x^2] = \dfrac{8x}{2\sqrt{9+4x^2}}

Otherwise if you don't like that method, make trig/hyperbolic substitutions in those situations - I would suggest trying x=32sinh⁡t x = \frac{3}{2} \sinh t in this case.


I would have thought that u=9+4x2u=9+4x^2 would be a nicer substitution...
Reply 5
Avatar for Josephiiine
Josephiiine
OP
spread_logic_not_hate
It's done on sight as it is presence of the derivative... consider that

ddx[9+4x2]=129+4x2×ddx[4x2]=8x29+4x2 \dfrac{d}{dx}[\sqrt{9+4x^2}] = \dfrac{1}{2\sqrt{9+4x^2}} \times \dfrac{d}{dx}[4x^2] = \dfrac{8x}{2\sqrt{9+4x^2}}

Otherwise if you don't like that method, make trig/hyperbolic substitutions in those situations - I would suggest trying x=32sinh⁡t x = \frac{3}{2} \sinh t in this case.





aaahh ok, yyeah the derivative thing makes sense, thank you! (: i would never have thought of that
meatball893
I would have thought that u=9+4x2u=9+4x^2 would be a nicer substitution...


True, didn't consider that!
Josephiiine
aaahh ok, yyeah the derivative thing makes sense, thank you! (: i would never have thought of that


When integrating stuff like that, look to see if anything in the denominator will differentiate to give you something in the numerator - you'll find you'll spot them more often then! For example

∫sin⁡xcos⁡2xdx \displaystyle \int \dfrac{\sin x}{\cos^2 x} dx

I see that the denominator will differentiate to give me the numerator, hence the answer is

1cos⁡x \dfrac{1}{\cos x}

Related discussions

Latest

Trending

Trending

Articles for you