# Integrals Rp_4-5

• Jan 6th 2010, 12:10 PM
gilyos
Integrals Rp_4-5
$\displaystyle \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx=$

$\displaystyle \int 5^{\sqrt{7x+2}}dx=$

In the secound problem i tried maybe $\displaystyle e^{(\sqrt{7x+2}*ln{5})}$ but I don't know how to continue...
• Jan 6th 2010, 04:14 PM
DeMath
Quote:

Originally Posted by gilyos
$\displaystyle \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx=$

$\displaystyle \int {\frac{{dx}} {{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}} = \int {\frac{{dx}} {{{{\left( {{{\left( {x - 1} \right)}^2} + 4} \right)}^{3/2}}}}} = \left\{ \begin{gathered} x - 1 = 2\tan u, \hfill \\ dx = \frac{{2du}} {{{{\cos }^2}u}} \hfill \\ \end{gathered} \right\} =$

$\displaystyle = \frac{1} {4}\int {\frac{{du}} {{{{\cos }^2}u{{\left( {{{\tan }^2}u + 1} \right)}^{3/2}}}}} = \frac{1} {4}\int {\frac{{du}} {{{{\cos }^2}u{{\left( {\frac{1} {{{{\cos }^2}u}}} \right)}^{3/2}}}}} = \frac{1} {4}\int {\frac{{{{\cos }^3}u}} {{{{\cos }^2}u}}\,du} =$

$\displaystyle = \frac{1} {4}\int {\cos u\,du} = \frac{1} {4}\sin u + C = \frac{1} {4}\sin \arctan \frac{{x - 1}} {2} + C =$

$\displaystyle = \frac{1} {4}\frac{{\frac{{x - 1}} {2}}} {{\sqrt {1 + {{\left( {\frac{{x - 1}} {2}} \right)}^2}} }} + C = \frac{{x - 1}} {{4\sqrt {{x^2} - 2x + 5} }} + C.$

Finally

$\displaystyle \int\limits_1^2 {\frac{{dx}} {{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}} = \left. {\frac{{x - 1}} {{4\sqrt {{x^2} - 2x + 5} }}} \right|_1^2 = \frac{1} {{4\sqrt 5 }} = \frac{{\sqrt 5 }} {{20}}.$
• Jan 6th 2010, 04:36 PM
DeMath
Quote:

Originally Posted by gilyos
$\displaystyle \int 5^{\sqrt{7x+2}}dx=$

In the secound problem i tried maybe $\displaystyle e^{(\sqrt{7x+2}*ln{5})}$ but I don't know how to continue...

Try this substitution $\displaystyle 7x + 2 = u^2$, after the integration by parts:

$\displaystyle \int {{5^{\sqrt {7x + 2} }}\,dx} = \left\{ \begin{gathered} 7x + 2 = {u^2} \hfill \\ dx = \frac{{2u}} {7}\,du \hfill \\ \end{gathered} \right\} = \frac{2} {7}\int {{5^u}u\,du} = \frac{{2 \cdot {5^u}u}} {{7\ln 5}} - \frac{2} {{7\ln 5}}\int {{5^u}\,du} =$

$\displaystyle = \frac{{2 \cdot {5^u}u}} {{7\ln 5}} - \frac{{2 \cdot {5^u}}} {{7{{\ln }^2}5}} + C = \left( {\frac{{2u}} {{7\ln 5}} - \frac{2} {{7{{\ln }^2}5}}} \right){5^u} + C =$

$\displaystyle = \left( {\frac{{2\sqrt {7x + 2} }} {{7\ln 5}} - \frac{2} {{7{{\ln }^2}5}}} \right){5^{\sqrt {7x + 2} }} + C.$
• Jan 7th 2010, 11:39 PM
gilyos
$\displaystyle = \frac{1} {4}\int {\cos u\,du} = \frac{1} {4}\sin u + C = \frac{1} {4}\sin \arctan \frac{{x - 1}} {2} + C =$

How you transposed this ? from sin(arctan)

to this :

$\displaystyle = \frac{1} {4}\frac{{\frac{{x - 1}} {2}}} {{\sqrt {1 + {{\left( {\frac{{x - 1}} {2}} \right)}^2}} }} + C = \frac{{x - 1}} {{4\sqrt {{x^2} - 2x + 5} }} + C.$
• Jan 7th 2010, 11:44 PM
Prove It
Quote:

Originally Posted by gilyos
$\displaystyle = \frac{1} {4}\int {\cos u\,du} = \frac{1} {4}\sin u + C = \frac{1} {4}\sin \arctan \frac{{x - 1}} {2} + C =$

How you transposed this ? from sin(arctan)

to this :

$\displaystyle = \frac{1} {4}\frac{{\frac{{x - 1}} {2}}} {{\sqrt {1 + {{\left( {\frac{{x - 1}} {2}} \right)}^2}} }} + C = \frac{{x - 1}} {{4\sqrt {{x^2} - 2x + 5} }} + C.$

$\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$.
• Jan 7th 2010, 11:45 PM
Chris L T521
Quote:

Originally Posted by gilyos
$\displaystyle = \frac{1} {4}\int {\cos u\,du} = \frac{1} {4}\sin u + C = \frac{1} {4}\sin \arctan \frac{{x - 1}} {2} + C =$

How you transposed this ? from sin(arctan)

to this :

$\displaystyle = \frac{1} {4}\frac{{\frac{{x - 1}} {2}}} {{\sqrt {1 + {{\left( {\frac{{x - 1}} {2}} \right)}^2}} }} + C = \frac{{x - 1}} {{4\sqrt {{x^2} - 2x + 5} }} + C.$

Treat $\displaystyle \vartheta=\arctan\frac{x-1}{2}$. Construct a triangle from this and then find $\displaystyle \sin\vartheta$.

You should then see that $\displaystyle \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}}$ as DeMath stated.
• Jan 8th 2010, 12:23 AM
gilyos
Quote:

Originally Posted by Chris L T521
Treat $\displaystyle \vartheta=\arctan\frac{x-1}{2}$. Construct a triangle from this and then find $\displaystyle \sin\vartheta$.

You should then see that $\displaystyle \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}}$ as DeMath stated.

So , if i get it right

$\displaystyle arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}$

?
• Jan 8th 2010, 01:13 AM
Prove It
Quote:

Originally Posted by gilyos
So , if i get it right

$\displaystyle arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}$

?

See my post:

$\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$.

So if $\displaystyle \theta = \arctan{\frac{x - 1}{2}}$ then

$\displaystyle \sin{\arctan{\frac{x - 1}{2}}} = \frac{\tan{\arctan{\frac{x - 1}{2}}}}{\sqrt{1 + \left(\tan{\arctan{\frac{x - 1}{2}}}\right)^2}}$

$\displaystyle = \frac{\frac{x - 1}{2}}{\sqrt{1 + \left(\frac{x - 1}{2}\right)^2}}$

$\displaystyle = \frac{\frac{x - 1}{2}}{\sqrt{\frac{4 + (x - 1)^2}{4}}}$

$\displaystyle = \frac{\frac{x - 1}{2}}{\frac{\sqrt{4 + x^2 - 2x + 1}}{2}}$

$\displaystyle = \frac{x - 1}{\sqrt{x^2 - 2x + 5}}$.