# Math Help - Integrals, what did I do wrong?

1. ## Integrals, what did I do wrong?

Kind of Frustrated at this, finals tomorrow and I'm reviewing the last concept for the day, integrals.

Interal of sin(x)cos(x)dx with upper bound of pi/4, and lower bound 0.

I let u=sinx
and du=cos(x)dx

Which turned it into u du and then I integrated it for u^2/2. Which becomes (sin(x))^2/2 . Sin of 0 will just be 0 so I didn't worry about that bound, so I plug in pi/4 and get 1/2 / 2 = 1/4. But that's wrong. What's my mistake?

2. Sinpi/4 = sqrt2 / 2 . Which when squared will equal 2/4 = 1/2. Is that wrong?

3. $\int_{0}^{\frac{\pi}{4}}\sin x \cos x~dx = \frac{1}{4}$

5. Originally Posted by Hibijibi
Kind of Frustrated at this, finals tomorrow and I'm reviewing the last concept for the day, integrals.

Interal of sin(x)cos(x)dx with upper bound of pi/4, and lower bound 0.

I let u=sinx
and du=cos(x)dx

Which turned it into u du and then I integrated it for u^2/2. Which becomes (sin(x))^2/2 . Sin of 0 will just be 0 so I didn't worry about that bound, so I plug in pi/4 and get 1/2 / 2 = 1/4. But that's wrong. What's my mistake?

Here is another way to do it

$\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$

$\int_{0}^{\pi/4}\sin(x)\cos(x)dx=\frac{1}{2}\int_{0}^{\pi/4}\sin(2x)dx=-\frac{1}{4}\cos(2x)|_{0}^{\pi/4}=-\frac{1}{4}\left( 0-1\right)=\frac{1}{4}$

6. Alright thanks. For the intergral of sin(2x) dx, what's the basic first step I should take? Turn it into something?

7. Originally Posted by Hibijibi
Alright thanks. For the intergral of sin(2x) dx, what's the basic first step I should take? Turn it into something?
$\int\sin(2x)dx$

let u =2x du=2dx

$\int\sin(2x)dx=\frac{1}{2}\int\sin(u)du=-\frac{1}{2}\cos(u)+c=-\frac{1}{2}\cos(2x)+c$

8. Thanks! This forum is a huge help considering I can't go to the math center at the college on weekends.

9. Originally Posted by Hibijibi
Thanks! This forum is a huge help considering I can't go to the math center at the college on weekends.

Yeah! I love this place.

10. Instead of making a new thread I thought I'd just post another problem here, keeps the clutter down.

Integral of (x^2-1)/x dx

What should u be and I'm just curious on to why you know what it should be? (I'm more interested in the conceptual thinking of the problem more than the actual steps to the solution.

11. Originally Posted by Hibijibi
Instead of making a new thread I thought I'd just post another problem here, keeps the clutter down.

Integral of (x^2-1)/x dx

What should u be and I'm just curious on to why you know what it should be? (I'm more interested in the conceptual thinking of the problem more than the actual steps to the solution.
No u sub is needed. Simplfy the integrand by using termwise division

Example: $\frac{3x^3+x-2}{x}=3x^2+1-\frac{2}{x}$

Good luck.

12. Need a little more description on what just happened there. You divided the numerator by the denominator? x^2-1 divided by x?

Edit, never mind I see it. Thanks : )

Integral of x/sqrt(1+3x^2) dx

14. Originally Posted by Hibijibi

Integral of x/sqrt(1+3x^2) dx
let u=the radicand or $u=1+3x^2$

Why would we pick that?

15. You pick it because when you take the derivative for du you can get an x out of it? Which is what you need to put all terms into u? But then what's the next step?

If u= 3x^2+1
du= 6x dx

You can counter balance the 6 by putting 1/6 outside the integral.

So now we have du/sqrt(u). And I guess we could change this to du u^(-1/2), and then integrate to be (1/6)(2)u^(1/2). And from there it's easy, but did I do everything up to there correctly? Or since it's 1/sqrtu, I could do ln|sqrt(u)|???

Edit again: I'm getting the wrong answer. My finished integral was 1/3*sqrt(3x^2+1) with bounds from -3 to 1. Plug in and I got 2/3 - 2/3*sqrt(7), which is wrong.

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