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[SOLVED] Integral in the form of 1\f(x)

karush

Well-known member
Jan 31, 2012
2,771
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$

my first step with this was

$\displaystyle4\int_0^\frac{1}{2}\frac{1}{1+4t^2}dt$

thot this could be a log rule. but doesn't seem to fit into that

the answer to this is $\frac{\pi}{2}$
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$

my first step with this was

$\displaystyle4\int_0^\frac{1}{2}\frac{1}{1+4t^2}dt$

thot this could be a log rule. but doesn't seem to fit into that

the answer to this is $\frac{\pi}{2}$
Note that

\[\int_0^{\frac{1}{2}}\frac{4}{1+4t^2}\,dt = 4\int_0^{\frac{1}{2}}\frac{1}{1+(2t)^2}\,dt\]

Make the substitution $u=2t$ to get

\[2\int_0^1\frac{1}{1+u^2}\,du\]

This should integrate to something familiar.

I hope this helps!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Just a note on top of what Chris L T521 wrote, the solution will not contain a logarithm. It requires trig substitution which you might not have seen yet. I don't want you to keep pursuing the idea on this that the answer will contain a logarithm.
 

karush

Well-known member
Jan 31, 2012
2,771
ok saw using trig substitution...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
As a further hint, observe that if we let:

$\displaystyle u=\tan(v)$

then:

$\displaystyle du=\sec^2(v)\,dv=(1+\tan^2(v))\,dv=(1+u^2)\,dv\, \therefore \,dv=\frac{1}{1+u^2}\,du$

Now, rewrite the definite integral in terms of v, making sure to also rewrite the limits of integration in terms of v.
 

karush

Well-known member
Jan 31, 2012
2,771
presume we are referring to the base form of.

$\displaystyle \int\frac{dx}{x^2+a^2}=\frac{1}{a}tan^{-1}\left(\frac{x}{a}\right)$

BTW why is there a 1 on the interval
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
BTW why is there a 1 on the interval
Chris made a substitution from the variable $t$ to the variable $u$. Since these are not the same, the bounds won't stay the same. If the interval for variable $t$ is $[0,1/2]$ and $u=2t$ then the interval in terms of $u$ is $[2*0,2*1/2]$ or simply $[0,1]$.

When doing substitution problems you can either rewrite the interval in terms of the new variable or solve the indefinite integral and then rewrite everything in terms of the first variable. Just make sure you apply the correct bounds.