Unsolved Integral Questions: Methods and Attempts

[transgalactic]

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

i could'nt solve them please help

 

[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 .

 

[Flux = Rad]

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.

 

[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

 

[transgalactic]

thank you very much