# 2 integral questions

1. Apr 20, 2007

### transgalactic

i added a file with the the two integral and the method that i tried to solve them

i could'nt solve them plz help

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• ###### i5.GIF
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Last edited: Apr 20, 2007
2. Apr 20, 2007

### Gib Z

Your substitutions seem to be the way to go for the first one. I hope you know those expressions for sin and cos actually come about from t= tan(x/2)?

Therefore by the chain rule, $dt=\frac{1}{2} \sec^2 (x/2) dx$.
Then we use the Pythatgorean Identities to reduce sec^2 to tan^2 +1.

$$dt = \frac{1}{2} (\tan^2 (x/2) +1 ) dx = \frac{1}{2} (t^2 +1) dx$$
Getting dx alone yields $$\frac{2 dt}{t^2+1} = dx$$ which unfortunately is not what you had. So try it with that correction.

For the second one, in partial fractions the numerator always has to be 1 degree less than the denominator. I sure you knew that because for the first partial fraction you put Ax+b over 1+x^2. The second partial fraction has degree 4, so instead of putting cx+d, try cx^3+dx^2+e^x+f .

3. Apr 20, 2007

For the second one, instead of splitting it into partial fractions try the substitution:
$$x = tan(u)$$

There's still a fair bit of work to do after that, but it should get you started.

4. Apr 20, 2007

### Gib Z

Actually, for the second one there is a much easier way that I just saw!

x= tan z
dx = sec^2 z dz
$$\int \frac{1}{(1+x^2)^2} dx = \int \frac{\sec^2 z}{(1+\tan^2 z)^2} dz=\int \frac{\sec^2 z}{\sec^4 z} dz =\int \frac{1}{\sec^2 z}dz = \int \cos^2 z dz$$

Now finish that off with $\cos^2 z = \frac{1}{2} (\cos 2z +1)$

EDIT: DAMN IT

5. Apr 21, 2007

### transgalactic

thank you very much