# Math Help - Substitution method for antiderivatives.

1. ## Substitution method for antiderivatives.

Tried my best with Latex, took forever!

1) $\int12x(6x^2+1)^3dx$
Let u= $6x^2+1$
$\frac {du}{dx}=12x$
$du=12xdx$

$\int(u)^312xdx= \int(u)^3du= \frac {u^4}{4} + C$

= $\frac {1}{4}(6x^2+1)^4 + C$

2) $\int \frac {3x^2dx}{\sqrt[3]{x^3+1}}$

Let u= $(x^3+1)$
$du= 3x^2dx$

$\int \frac {du}{\sqrt[3]{u}}$ = $u^{-1/3}du$
= $\frac {u^{2/3}}{2/3} + C$

= $\frac {3}{2}(x^3+1)^{2/3} + C$

3)
$\int 2x\cos{(x^2)}dx$

Let u= $(x^2)$ du= $2xdx$
$
\int \cos{x^2}dx2x= \cos{(u)}du

=\sin{x^2} + C$

**Edit** One more

4) $\int \sin{x}^5 \cos{x}dx$

2. These look good. You can always find dy/dx of the answer to check them.

3. Awesome, thanks! I appreciate your help. I wish I could try more...I emailed my prof. and he said all of the ones in the book are too high a level and he hasn't taught it yet. Guess I'll just have to wait. Also, note that I added one more question.

4. I assume 4. is $\displaystyle \int{\sin^5{x}\cos{x}\,dx}$

Make the substitution $\displaystyle u = \sin{x}$.

5. Yeah, that's it. I did make that substitution. What did I do wrong? I thought sin^5x and sinx^5 were the same thing? Or am I seriously mistakin'?

6. If you write $\displaystyle \sin{x}^5$ we can't tell if it's $\displaystyle \sin{(x^5)}$ or $\displaystyle (\sin{x})^5$...

So after you make the substitution $\displaystyle u = \sin{x}$, what's $\displaystyle \frac{du}{dx}$? What's $\displaystyle du$?

7. Originally Posted by Prove It
If you write $\displaystyle \sin{x}^5$ we can't tell if it's $\displaystyle \sin{(x^5)}$ or $\displaystyle (\sin{x})^5$...

So after you make the substitution $\displaystyle u = \sin{x}$, what's $\displaystyle \frac{du}{dx}$? What's $\displaystyle du$?
cos(x)dx

8. So what is your new integral in terms of u?

9. Originally Posted by Prove It
So what is your new integral in terms of u?
(1/6) (sinx) ^6 + C.

Is this not the answer? All the work I'd give would leave to that.

10. $\displaystyle y= \frac{1}{6}\sin ^6x+C \implies \frac{dy}{dx} = \sin^5 x \cos x$

Looks good.

11. Sweet, thanks! Can't wait for the more challenging ones.

12. Originally Posted by Marconis
Sweet, thanks! Can't wait for the more challenging ones.
Try this one

$\displaystyle \int\frac{x+1}{x-1}~dx$

13. Originally Posted by pickslides
Try this one

$\displaystyle \int\frac{x+1}{x-1}~dx$
Seems like it's hard, lol. I'll get back to you tomorrow. Time to enjoy my Friday night.

14. Hey, so I didn't get a chance do the question you provided, had a busy weekend. I will post the following, though:

$\int \cos^4{x} \sin{x} dx$

$u= \cos{x}$
$du=- \sin{x}dx$

$-1 \int (u)^4 du$
$=-1 \int \frac {u^{5}}{5} + C$
$= \frac {-1}{5} (\cos{x})^5 + C$

What do you guys think?

15. Originally Posted by Marconis
Hey, so I didn't get a chance do the question you provided, had a busy weekend. I will post the following, though:

$\int \cos^4{x} \sin{x} dx$

$u= \cos{x}$
$du=- \sin{x}dx$

$-1 \int (u)^4 du$
$=-1 \int \frac {u^{5}}{5} + C$ Get rid of the integration sign here
$= \frac {-1}{5} (\cos{x})^5 + C$

What do you guys think?
With this one small change everything is fine

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