# Thread: do u wanna solve perfect ıntegration questions?:)

1. ## do u wanna solve perfect ıntegration questions?:)

1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????

2)INT dx/[x(1+x^3+x^6)^1/2]????????

3)INT (e^2x)/(x^2-3x+2)dx ???????

4)INT (ch^2)(sh^2)dx????????

5)INT (ch^4)dx ????????

2. Originally Posted by berkanatci
1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????

2)INT dx/[x(1+x^3+x^6)^1/2]????????

3)INT (e^2x)/(x^2-3x+2)dx ???????

4)INT (ch^2)(sh^2)dx????????

5)INT (ch^4)dx ????????

Make sure you write these accurately:

INT (arctane^x/2)/[e^x/2(1+e^x)]dx

Do you mean:
INT arctan(x/2)/[e^(x/2)*(1 + e^x)] dx ?

Go through all the problems that you need help on and make sure they are written accurately, otherwise I can't help you if I'm solving the wrong problem.

3. Originally Posted by ecMathGeek
Make sure you write these accurately:

INT (arctane^x/2)/[e^x/2(1+e^x)]dx

Do you mean:
INT arctan(x/2)/[e^(x/2)*(1 + e^x)] dx ?

Go through all the problems that you need help on and make sure they are written accurately, otherwise I can't help you if I'm solving the wrong problem.

the correct one is:INT [arctane^(x/2)]/[e^(x/2)*(1+e^x)]dx?

I wrote it correctly

4. Originally Posted by berkanatci
1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????
Help is a little late, but here's number 1. And please use parenthesis!

$\displaystyle \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx$

I'll transform the integral in steps:
Let $\displaystyle y = e^{x/2} \implies dy = \frac{e^{x/2}}{2} dx$

Thus
$\displaystyle \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx = \int atn(y) \frac{2dy}{1 + y^2} = 2 \int atn(y) \frac{dy}{1 + y^2}$

Let $\displaystyle z = atn(y) \implies dz = \frac{dy}{1 + y^2}$

Finally:
$\displaystyle \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx = 2 \int atn(y) \frac{dy}{1 + y^2} = 2 \int z dz = z^2 + C$

$\displaystyle = atn^2(y) + C = atn^2 \left ( e^{x/2} \right ) + C$

-Dan

5. 2) $\displaystyle \displaystyle\int\frac{dx}{x(1+x^3+x^6)^{\frac{1}{ 2}}}=\int\frac{x^2dx}{x^3(1+x^3+x^6)^{\frac{1}{2}} }$
Let $\displaystyle x^3=t\Rightarrow 3x^2dx=dt$
Then $\displaystyle I=\frac{1}{3}\int\frac{dt}{t\sqrt{t^2+t+1}}$
Now $\displaystyle \displaystyle t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$
$\displaystyle \displaystyle I=-\frac{1}{3}\int\frac{du}{\sqrt{u^2+u+1}}=-\frac{1}{3}\ln\left(u+\frac{1}{2}+\sqrt{u^2+u+1}\r ight)+C$
Now back substitute.

4) $\displaystyle I=\int ch^2xsh^2xdx=\displaystyle\frac{1}{4}\int sh2xdx=\frac{1}{8}ch2x+C$

5) $\displaystyle \int ch^4xdx=\int ch^2x(1+sh^2x)dx=\int ch^2xdx+\int ch^2xsh^2xdx=$
$\displaystyle \displaystyle =\frac{1}{2}\int(1+ch2x)dx+\frac{1}{8}ch2x=\frac{x }{2}+\frac{1}{4}sh2x+\frac{1}{8}ch2x+C$

6. Originally Posted by red_dog
2) $\displaystyle \displaystyle\int\frac{dx}{x(1+x^3+x^6)^{\frac{1}{ 2}}}=\int\frac{x^2dx}{x^3(1+x^3+x^6)^{\frac{1}{2}} }$
Let $\displaystyle x^3=t\Rightarrow 3x^2dx=dt$
Then $\displaystyle I=\frac{1}{3}\int\frac{dt}{t\sqrt{t^2+t+1}}$
Now $\displaystyle \displaystyle t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$
$\displaystyle \displaystyle I=-\frac{1}{3}\int\frac{du}{\sqrt{u^2+u+1}}=-\frac{1}{3}\ln\left(u+\frac{1}{2}+\sqrt{u^2+u+1}\r ight)+C$
Now back substitute.
Oh that's just too cute!

-Dan

7. Originally Posted by topsquark
Oh that's just too cute!

-Dan
Completely agree. excellent work red_dog!

8. Originally Posted by berkanatci
5)INT (ch^4)dx ????????
I assume it means,
$\displaystyle \int \cosh^4 x dx$

You can use the identity, $\displaystyle \cosh^2 x = \frac{\cosh 2x + 1}{2}$ twice.

4)INT (ch^2)(sh^2)dx????????
$\displaystyle \int \cosh^2 x \sinh^2 x dx$
Use "half angle identiy":
$\displaystyle \int \left( \frac{\cosh 2x + 1}{2} \right) \left( \frac{\cosh 2x - 1}{2} \right) dx$

The details are easy.

9. Originally Posted by berkanatci
3)INT (e^2x)/(x^2-3x+2)dx ???????
Write,
$\displaystyle \int \frac{e^{2x}}{(x-2)} - \frac{e^{2x}}{(x-1)} dx$

Consider the first integral,
$\displaystyle \int \frac{e^{2x}}{x-2} dx$ let $\displaystyle t = x-2 \implies t'=1$.
Thus,
$\displaystyle \int \frac{e^{2t}e^2}{t} dt = \int \frac{2e^{2t}e^2}{2t} dt = e^2\mbox{Ei}(2t)+C = e^2\mbox{Ei}(2x-4)+C$

Similarly do the other one.