1. ## Integrals

Hi,

Q1. Evaluate the following integrals by suing the given substitution.

a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

b) f { x*sin(x^2) } dx, substitution u = x^2

c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

(*note: the functions above are antiderivatives...)

can anyone help me with these questions? im clueless - steps and answers would be appreciated...

thanks,

moon

2. Originally Posted by Moon Hoplite
Hi,

Q1. Evaluate the following integrals by suing the given substitution.

a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

b) f { x*sin(x^2) } dx, substitution u = x^2

c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

(*note: the functions above are antiderivatives...)

can anyone help me with these questions? im clueless - steps and answers would be appreciated...

thanks,

moon
a) $\int{\frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx}$.

Notice that this can be rewritten as $\int{e^{\sqrt{x}}\frac{1}{2\sqrt{x}}\,dx}$.

Like you were told, make the substitution $u = \sqrt{x} = x^{\frac{1}{2}}$.

$\frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$.

So the integral reduces to

$\int{e^u\,\frac{du}{dx}\,dx}$

$= \int{e^u\,du}$

$= e^u + C$

$= e^{\sqrt{x}} + C$.

3. Originally Posted by Moon Hoplite
Hi,

Q1. Evaluate the following integrals by suing the given substitution.

a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

b) f { x*sin(x^2) } dx, substitution u = x^2

c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

(*note: the functions above are antiderivatives...)

can anyone help me with these questions? im clueless - steps and answers would be appreciated...

thanks,

moon
$\int{x\sin{x^2}\,dx}$.

Use the substitution $u = x^2$.

$\frac{du}{dx} = 2x$.

We need to rearrange the integral so it has $2x$ as a factor. Notice that $x = \frac{1}{2}\times 2x$

So we can rewrite this as

$\frac{1}{2}\int{2x\sin{x^2}\,dx}$

$= \frac{1}{2}\int{\sin{u}\,\frac{du}{dx}\,dx}$

$= \frac{1}{2}\int{\sin{u}\,du}$

$= -\frac{1}{2}\cos{u} + C$

$= -\frac{1}{2}\cos{x^2} + C$.

4. Originally Posted by Moon Hoplite
Hi,

Q1. Evaluate the following integrals by suing the given substitution.

a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

b) f { x*sin(x^2) } dx, substitution u = x^2

c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

(*note: the functions above are antiderivatives...)

can anyone help me with these questions? im clueless - steps and answers would be appreciated...

thanks,

moon
c) $\int{\frac{2x + 5}{x^2 + 5x + 17}\,dx}$.

First, rewrite this as

$\int{\frac{1}{x^2 + 5x + 17}(2x + 5)\,dx}$.

Make the substitution $u = x^2 + 5x + 17$ so that $\frac{du}{dx} = 2x + 5$.

So the integral becomes

$\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$= \int{\frac{1}{u}\,du}$

$= \ln{|u|} + C$

$= \ln{|x^2 + 5x + 17|} + C$.