Space between these functions of x

[Dell]

i need to find the are trapped between the graphs of
y=-x³
y=arctg\sqrt{x}
x=2
x=1

y=arctg\sqrt{x} will be my top function so the area will be

\intarctg\sqrt{x}-(-x³)dx from 1 to 2

in order to to this i can split the integral up and get
\intarctg\sqrt{x}dx -\int-x³dx both 1-2

|\int-x³dx| (1-2) is 3.75 with a simple table integral

but i don't know how to integrate the arctg

what i think could work is...

arctg\sqrt{x}=t
tg(t)=\sqrt{x}

\inttdx while dt=dx/1+x²

\intt(1+x⁴)dt
=\intt(1+tg⁴t)dt
=\inttdt+\int(tg⁴t)dt

another option
\intt(1+x⁴)dt
=1/2\int(1+tg⁴t)dt²
=0.5\intdt²+0.5\int(tg⁴t)dt²

sort of feels like I am getting lost in it around here,,, any ideas??

 

[Dell]

how about this,,

lets call t²=p

\int(tg²p)dp=\int\frac{sin^2p}{cos^2p}

=\int(1/cos²p)dp-\intdp
=tgp-p
=tg(t²)-t²
=tg[{arctg\sqrt{x}}²]-(arctg\sqrt{x})²
=x-(arctg\sqrt{x})²

is this correct

 

[Dick]

To integrate arctg(sqrt(x)), set u=sqrt(x). So du=dx/(2*sqrt(x))=dx/(2u). Or 2udu=dx. The integral in terms of u becomes arctg(u)*2udu=arctg(u)*d(u^2). Now integrate by parts.

 

[Dell]

on my table of integrals i don't have an integral for arctg, how do i do this? do i need to do something to the effect of

arctg u =v
tgv=u and then carry on from there (similar to what i posted? ) or is there a abetter way?

 

[Dick]

To integrate arctg(u)*d(u^2) you don't need to know how to integrate arctg(u). You only need to know how to differentiate it. a*db=d(a*b)-b*da. That's how integration by parts works, right?

 

[Dell]

cant see it, could you please show me.

 

[Dick]

The integral of arctg(u)*d(u^2)=arctg(u)*u^2-the integral of u^2*d(arctg(u)). That's what integration by parts means, yes? What d(arctg(u))? Differentiate arctg(u).

 

[Dell]

ive read up a bit on this and found an equation

\intudv=u*v-\intvdu

is this what you mean??

so in my case
arctgU is my u function
and u^2 is my v function
where u is the root of x

so

\int(arctg(u))d(u²)=(arctg(u))*u²-\intu²d(arctg(u))

now d(arctg(u))=\frac{1}{1+u^2}*d(u^2)

so \intu²d(arctg(u))=\intU²/(1+u²)d(u²⁾

correct up till here??

=\int1-\frac{1}{1+u^2}du²====u²-ln|1+u^2|

which is x-ln|a+x|

 

[Dick]

You are getting there. But d(arctg(u))=du/(1+u^2) (NOT d(u^2)). So the integral you are left with is u^2*du/(1+u^2). You'll find there aren't any logs in the problem.

 

[Dell]

how did you come to d(arctg(u))=du/(1+u^2)
is (in general) d(t) not equal to the differential of t multiplied by dx?

in this case d(arctg(u))= differential of (arctg(u)) * d(u^2)

how do i know which d() to multiply by?

 

[Dick]

how did you come to d(arctg(u))=du/(1+u^2)
is (in general) d(t) not equal to the differential of t multiplied by dx?

in this case d(arctg(u))= differential of (arctg(u)) * d(u^2)

how do i know which d() to multiply by?

Exactly as you said. d(f(x))=f'(x)*dx. d(arctg(x))=(d/dx arctg(x))*dx. d/dx(arctg(x))=1/(1+x^2). So d(arctg(x))=dx/(1+x^2). I don't know where you are getting the d(u^2) stuff.

 

[Dell]

i see my problem, what is the differential of arctg(x^0.5) is it not the 1/(1+x^0.5) multiplied by 1/2(x^0.5) ??

 

[Gib Z]

Well, it appears you are using the chain rule, which is good, but you are forgetting to square something. Look at what the derivative of arctan x is again.

 

[Dick]

i see my problem, what is the differential of arctg(x^0.5) is it not the 1/(1+x^0.5) multiplied by 1/2(x^0.5) ??

No! The differential of arctg(sqrt(x)) is 1/(1+sqrt(x)^2) times 1//(2*sqrt(x)). Like Gib said, you forgot the square in the derivative of arctg. That's EXACTLY the same thing as saying the differential of arctg(u) is 1/(1+u^2) times du where u=sqrt(x). Look at it!