subsititution

[xc630]

Hello I would appreciate some help with a few problems

1)

if the substitution u= SQRT (x-1) is made, the integral from 2 to 5 of SQRT(x-1)/x dx = ?

I solved du to = 1/ 2SQRT (x-1) dx but I can't subistute that for the 1/x in the denominator of the original equation. how would I manipulate it to match?

2) I need to simplify the integral of (sec^2 (x)) (tan^2(x)) dx
I tried letting u = sec^2(x)) but then du would equal tan x dx. That wouldn't substitute well either though.

 

[galactus]

\(\displaystyle \L\\\int_{2}^{5}\frac{\sqrt{x-1}}{x}dx\)

As you have done, let \(\displaystyle \L\\u=\sqrt{x-1}, \;\ u^{2}+1=x, \;\ dx=2udu\)

\(\displaystyle \L\\2\int_{1}^{2}\frac{u^{2}}{u^{2}+1}du\)

\(\displaystyle \L\\=\int_{1}^{2}2du-\int_{1}^{2}\frac{2}{u^{2}+1}du\)

For #2, simply let \(\displaystyle \L\\u=tan(x), \;\ du=sec^{2}(x)dx\)

 

[xc630]

I do not understand how the interval changed from 2 to 5 to 1 to 2. Also if you substitute wouldn't it be u and not u^2 in the numerator?

 

[galactus]

The limits changed because we changed variables. The extra u in the numerator comes from 2udu.

\(\displaystyle \L\\\int\frac{u}{u^{2}+1}\cdot{2udu}\)

Now, you have to chage the limits.

x=2 and 5 in the original.

\(\displaystyle \L\\u=\sqrt{(5)-1}=2\)

\(\displaystyle \L\\u=\sqrt{(2)-1}=1\)

See now?.

 

[xc630]

oh ok thank you