[SOLVED]Integral with substitution

karush

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

$\displaystyle U=\sqrt{t}\ \ \ t=U^2 \ \ \ dt=2Udu$

$\displaystyle \frac{\sqrt{t}}{t+1} \Rightarrow \frac{U}{U^2+1}$

$\displaystyle \int_0^4 \frac{U}{U^2+1} 2Udu$

if ok so far tried $U= sec^2(\theta)$

but couldn't not get answer which is

Krizalid

Active member
When you perform a substitution with a definite integral, you need to change the bounds. So for $0\le t\le4$ you get $0\le u\le 2$ since $u=\sqrt t.$ Now, note that $\dfrac{{{u}^{2}}}{{{u}^{2}}+1}=\dfrac{{{u}^{2}}+1-1}{{{u}^{2}}+1}=1-\dfrac{1}{{{u}^{2}}+1},$ so if you integrate the right side, you'll see immediate integrals there.

karush

Well-known member
When you perform a substitution with a definite integral, you need to change the bounds. So for $0\le t\le4$ you get $0\le u\le 2$ since $u=\sqrt t.$ Now, note that $\dfrac{{{u}^{2}}}{{{u}^{2}}+1}=\dfrac{{{u}^{2}}+1-1}{{{u}^{2}}+1}=1-\dfrac{1}{{{u}^{2}}+1},$ so if you integrate the right side, you'll see immediate integrals there.
So this will go to
$2\left[U-\tan^{-1}(U)\right]_0^4$
$2\left[√t-\tan^{-1}√t\right]_0^4=1.7857$

Last edited:

MarkFL

Staff member
You have the correct result, but you could have just used:

$$\displaystyle 2\left[u-\tan^{-1}(u) \right]_0^2=2\left(2-\tan^{-1}(2) \right)\approx1.78570256441182$$

Once you make a substitution in a definite integral and have rewritten the integrand, differential and limits in terms of the new variable, there is no need to consider the old variable again.