# direct integration

• Nov 13th 2012, 02:35 PM
Chipset3600
direct integration

$\displaystyle \int tan(x/2).sec^2(x/2)+sec^2(x/2)$
$\displaystyle t=(x/2)$
$\displaystyle \int tan(t). sec^2(t)dt+tan(t) = -ln[cos(t)]+tan(t)+C$

But isnt correct, i'm afraid my substitution is not correct
• Nov 13th 2012, 03:40 PM
MarkFL
Re: direct integration

$\displaystyle \int \tan\left(\frac{x}{2} \right)\sec^2\left(\frac{x}{2} \right)+\sec^2\left(\frac{x}{2} \right)\,dx$

I would first factor the integrand:

$\displaystyle \int \sec^2\left(\frac{x}{2} \right)\left(\tan\left(\frac{x}{2} \right)+1 \right)\,dx$

Now, let:

$\displaystyle u=\tan\left(\frac{x}{2} \right)+1\,\therefore\,du=\frac{1}{2}\sec^2\left( \frac{x}{2} \right)\,dx$

and the integral becomes:

$\displaystyle 2\int u\,du$

Now, find the anti-derivative, then back-substitute for $\displaystyle u$.
• Nov 13th 2012, 03:43 PM
Chipset3600
Re: direct integration
Quote:

Originally Posted by MarkFL2

$\displaystyle \int \tan\left(\frac{x}{2} \right)\sec^2\left(\frac{x}{2} \right)+\sec^2\left(\frac{x}{2} \right)\,dx$

I would first factor the integrand:

$\displaystyle \int \sec^2\left(\frac{x}{2} \right)\left(\tan\left(\frac{x}{2} \right)+1 \right)\,dx$

Now, let:

$\displaystyle u=\tan\left(\frac{x}{2} \right)+1\,\therefore\,du=\frac{1}{2}\sec^2\left( \frac{x}{2} \right)\,dx$

and the integral becomes:

$\displaystyle 2\int u\,du$

Now, find the anti-derivative, then back-substitute for $\displaystyle u$.

MarkFL i'm learning now integral in school and we dont use substitution yet(integration techniques)
.
• Nov 13th 2012, 03:47 PM
MarkFL
Re: direct integration
Quote:

Originally Posted by Chipset3600
...i'm afraid my substitution is not correct

Quote:

Originally Posted by Chipset3600
...we dont use substitution yet...

Do you see why I assumed you wanted to use a substitution?
• Nov 13th 2012, 03:51 PM
Chipset3600
Re: direct integration
Quote:

Originally Posted by MarkFL2
Do you see why I assumed you wanted to use a substitution?

Yes, sorry! I did it just to ease my calculations
• Nov 13th 2012, 09:00 PM
hollywood
Re: direct integration
This is a difficult integral for someone who hasn't learned about u-substitution yet. You're probably aware that the derivative of $\displaystyle \tan{\frac{x}{2}}$ is $\displaystyle \frac{1}{2}\sec^2{\frac{x}{2}}$. And looking at the other term might eventually lead you to take the derivative of $\displaystyle \tan^2{\frac{x}{2}}$. Once you do that, you should be able to combine the two derivatives to get the expression you want to integrate.

It's probably less work to first learn u-substution, then apply it to this integral. But I guess that's not an option for you.

- Hollywood
• Nov 15th 2012, 01:29 PM
Chipset3600
Re: direct integration
Quote:

Originally Posted by hollywood
This is a difficult integral for someone who hasn't learned about u-substitution yet. You're probably aware that the derivative of $\displaystyle \tan{\frac{x}{2}}$ is $\displaystyle \frac{1}{2}\sec^2{\frac{x}{2}}$. And looking at the other term might eventually lead you to take the derivative of $\displaystyle \tan^2{\frac{x}{2}}$. Once you do that, you should be able to combine the two derivatives to get the expression you want to integrate.

It's probably less work to first learn u-substution, then apply it to this integral. But I guess that's not an option for you.

- Hollywood

Actually isnt so hard, take a look:

$\displaystyle \int tan(x/2)sec^2(x/2)dx.(\frac{1}{2}.\frac{2}{1}) + \int sec^2(x/2)dx.(\frac{1}{2}.\frac{2}{1})$

$\displaystyle 2\int tan(x/2).\frac{1}{2} sec^2(x/2)dx + 2\int sec^2(x/2). \frac{1}{2}dx$

Now i use the power rule for the first and the sec^2(u) rule for the second integral

$\displaystyle tan^2(x/2)+ 2tan(x/2)+C$
Now take a look of the derivative of my answer: Attachment 25734