1. ## Integral

Don't know how to work out the following integral.
Substituting cos x^5 = t does not seem to work.

2. This looks much harder than it is.

Let $u=x^5, du=5x^4dx$

Another way of solving for "du" is $\frac{1}{5}du=x^4dx$. But we don't want just x^4dx, we want 2x^4, so multiply by 2 on both sides to get $\frac{2}{5}du=2x^4$. Now we're getting somewhere.

So plug everything back in. Our new integral in terms of u is...

$\frac{2}{5} \int \frac{1}{\cos^2(u)}du$

Now what's $\frac{1}{\cos^2(x)}$?? $\sec^2(x)$.

See it now?

3. Originally Posted by totalnewbie
Don't know how to work out the following integral.
Substituting cos x^5 = t does not seem to work.
Let,
$u=x^5$ then $u'=5x^4$
Thus,
$\frac{2}{5} \int \frac{5x^4}{\cos^2 x^5}dx$
Is,
$\frac{2}{5} \int \sec^2u u' dx=\frac{2}{5}\tan u+C=\frac{2}{5}\tan (x^5)+C$

4. What is SEC ? Don't find that kind of word from the dictionary.

5. Should I substitute again ?
I am not able to get rid of x from denominator

6. Originally Posted by totalnewbie
What is SEC ? Don't find that kind of word from the dictionary.
Secant
----
Your other problem that you posted cannot be done.
$\int e^{x^2}dx$ which cannot be done.

7. Originally Posted by totalnewbie
Should I substitute again ?
I am not able to get rid of x from denominator
Hello,

your problem looks harder than it is actually:

$\int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{\frac{1}{x} \cdot (\ln(x)+2)^{-\frac{2}{5}}dx}$

let $u=\ln(x)+2$. Then
$\frac{du}{dx}=\frac{1}{x}$. Thus $du=\frac{1}{x} dx$.
Now plug in these terms into the original integral:

$\int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{u^{-\frac{2}{5}} du}$

I'll leave the rest for you.

(For confirmation only: $\frac{5}{3}(\ln(x)+2)^{\frac{3}{5}}$

EB

8. Originally Posted by earboth
Hello,

your problem looks harder than it is actually:

$\int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{\frac{1}{x} \cdot (\ln(x)+2)^{-\frac{2}{5}}dx}$

let $u=\ln(x)+2$. Then
$\frac{du}{dx}=\frac{1}{x}$. Thus $du=\frac{1}{x} dx$.
Now plug in these terms into the original integral:

$\int{\frac{dx}{x\cdot \sqrt[5]{(\ln(x)+2)^2}}}=\int{u^{-\frac{2}{5}} du}$

I'll leave the rest for you.

(For confirmation only: $\frac{5}{3}(\ln(x)+2)^{\frac{3}{5}}$

EB
The problem is that he has,
$\frac{1}{x^5}$ not $\frac{1}{x}$

9. Originally Posted by ThePerfectHacker
The problem is that he has,
$\frac{1}{x^5}$ not $\frac{1}{x}$
Hi TPH,

the image of the problem i can see shows a multiplication sign between the x and the 5th root. So I believe that I read it correctly.

EB

10. $\int \arctan{x}dx$

Any ideas ?

11. Originally Posted by totalnewbie
$\int \arctan{x}dx$

Any ideas ?
One thing I hate about integral is checking differenciability intervals and domains. Since I am lazy here was my approach.

$\int \tan^{-1} x dx$
Express as,
$\int (1+x^2) \cdot \frac{\tan^{-1} x}{1+x^2} dx$
Now,
$1+x^2=\sec ^2 (\tan^{-1} x)$
Thus,
$\int \sec^2 (\tan^{-1} x) \cdot \tan^{-1} x\cdot \frac{1}{1+x^2} dx$
Let,
$f(x)=x\sec^2x$
$g(x)=\tan^{-1} x$
$g'(x)=1/(1+x^2)$
Thus,
$\int (f \circ g)\cdot g' dx=F\circ g+C$
Where,
$F$ is an anti-derivative of $x\sec^2 x$
So the problem reduces to finding,
$\int x\sec^2 xdx$
Which should be doable by integration by parts.

12. $\int \frac{xdx}{1+x^2}=\frac{1}{2}\int\frac{d(1+x^2)}{1 +x^2}=\frac{1}{2}ln|1+x^2|+C$

Where does $\frac{1}{2}$ come from ?
Where does $d(1+x^2)$ come from ?
If $\int\frac{dx}{1+x^2}=arctanx+C$ how is it possible that my answer has $ln$ ?

13. Originally Posted by totalnewbie
$\int \frac{xdx}{1+x^2}=\frac{1}{2}\int\frac{d(1+x^2)}{1 +x^2}=\frac{1}{2}ln|1+x^2|+C$

Where does $\frac{1}{2}$ come from ?
Where does $d(1+x^2)$ come from ?
If $\int\frac{dx}{1+x^2}=arctanx+C$ how is it possible that my answer has $ln$ ?
Becuse you are substituting $x^2+1$.
Its derivative is $2x$. Note the 2. In the numerator you just have $x$. So to get the 2. You multiply and divide by 2 (that is 1/2). That is where it comes from.

14. I would start off with integration by parts:

$
\int {\arctan x} dx = x\arctan x - \int {\frac{x}{{1 + x^2 }}dx}
$

Now, the second integral is easy:

$
\int {\frac{x}{{1 + x^2 }}dx} = \frac{1}{2}\int {\frac{1}{{1 + x^2 }}d\left( {1 + x^2 } \right) = \frac{1}{2}\ln \left| {1 + x^2 } \right|} + C
$

So combined:

$
\int {\arctan x} dx = x\arctan x - \frac{1}{2}\ln \left| {1 + x^2 } \right| + C
$

15. Originally Posted by totalnewbie
$\int \arctan{x}dx$

Any ideas ?
try $\tan(u)=x$

RonL

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