Integrals

**Sally_Math** · February 14th 2010, 03:50 AM

Can someone plz help me evaluate these integrals?

integral(

e^x(3 + e^x)^8 )dx

integral(PI/6 PI/4) (sin(x)/(1+cos(x))^2)dx

**Nyrox** · February 14th 2010, 04:01 AM

In both cases you can find a primitive quite quickly. Observe that

$\frac{d}{dx}(3+e^x)=e^x$ ,

so the chain rule yields, for a well-behaved function $g$ :

$\frac{d}{dx}g(3+e^x)=e^xg'(3+e^x)$ .

In particular, choosing $g(x)=x^9$ :

$\frac{d}{dx}(3+e^x)^9=9e^x(3+e^x)^8$ .

Consequently

$\int e^x(3+e^x)^8\; dx=\frac{1}{9}(3+e^x)^9$ .

Try applying the same method to the second one

**e^(i*pi)** · February 14th 2010, 04:05 AM

Originally Posted by Sally_Math

Can someone plz help me evaluate these integrals?

integral(

e^x(3 + e^x)^8 )dx

integral(PI/6 PI/4) (sin(x)/(1+cos(x))^2)dx

$\int ^{\pi/6}_{\pi/4} \left[ \frac{\sin (x)}{(1+\cos (x))^2}\right]$

Let $u = 1+\cos(x)<br />$
$du = -\sin(x) dx$

$dx = -\frac{du}{\sin(x)}$

Change the limits using our substitution:

$u_2 = 1+\cos \left(\frac{\pi}{6}\right) = \frac{2+\sqrt3}{2}$

$u_1 = 1+ \cos \left(\frac{\pi}{4}\right) = \frac{2+\sqrt2}{2}$

$<br /> -\int ^{\frac{2+\sqrt3}{2}}_{\frac{2+\sqrt2}{2}} \left(\frac{du}{u^2}\right)$

Which should be easy enough to solve

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