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

#### karush

##### Well-known member
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$

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

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

Staff member
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
ok saw using trig substitution...

#### MarkFL

Staff member
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
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

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 $[0,1]$.