# Math Help - Integration Help Calculus

1. ## Integration Help Calculus

I am new to this forum..so hello everyone! I am a student at UCD and sadly they make me take Calculus, even though I am a Political Science major...So please help me, I do not know if I did this right....

Find the Integral

(x + cos x) sin xdx the intervals are 0 and pi/2

My answer: [(sin x )^2]/2 + C...which is then .5-0 and then .5

Is this right?

Thank you so much!

2. Sorry, do you mean $\int_0^{\pi/2}{(x+\cos{x})\sin{x}}dx$?

3. Yes, sorry about the confusion. I have no clue how to write the problems in that form.

4. So, you want to find:

$\int_0^{\frac{\pi}{2}} [sin(x)(x + cos(x))]dx$

You can go ahead and distribute the sin(x):

$\int_0^{\frac{\pi}{2}} [xsin(x) + sin(x)cos(x)]dx$

Now, you can split the integral:

$\int_0^{\frac{\pi}{2}} [xsin(x)]dx + \int_0^{\frac{\pi}{2}} [sin(x)cos(x)]dx$

Let's go ahead and do the second integral:

$\int_0^{\frac{\pi}{2}} [sin(x)cos(x)]dx$

This can be solved with simple substitution:

$u = sin(x)$

$du = cos(x)dx$

Now we redo the limits in terms of u:

$u(0) = sin(0) = 0$

$u\left(\frac{\pi}{2}\right) = sin\left(\frac{\pi}{2}\right) = 1$

Our new integral is:

$\int_0^1 [u]du = \frac{u^2}{2}]^1_0 = \frac{1}{2}$

Now, we have to go back to the first integral:

$\int_0^{\frac{\pi}{2}} [xsin(x)]dx$

We have to use parts:

$u = x$

$du = dx$

$dv = sin(x)dx$

$v = -cos(x)$

Now, we have a complete integral:

$[-xcos(x)]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} [-cos(x)]dx$

Since $cos\left(\frac{\pi}{2}\right) = 0$ the first term is gone, which leaves the second term, which is:

$\int_0^{\frac{\pi}{2}} [cos(x)]dx$ <--- Cancelled the negatives

$= [sin(x)]_0^{\frac{\pi}{2}} = sin\left(\frac{\pi}{2}\right) = 1$

So, the total is:

$1 + \frac{1}{2} = \frac{3}{2}$

5. Originally Posted by bloomsgal8
I am new to this forum..so hello everyone! I am a student at UCD and sadly they make me take Calculus, even though I am a Political Science major...So please help me, I do not know if I did this right....

Find the Integral

(x + cos x) sin xdx the intervals are 0 and pi/2

My answer: [(sin x )^2]/2 + C...which is then .5-0 and then .5

Is this right?

Thank you so much!
$\int_{0}^{\frac{\pi}{2}}{\left(x+cosx\right)}sinx\ ,dx$.

You can't make a direct substitution quite yet. Distribute the sine term throughout the parenthesis to get this:

$\int_{0}^{\frac{\pi}{2}}{\left(xsinx+sinxcosx\righ t)}\,dx$.

Recall that $2sinxcosx=sin2x. \therefore sinxcosx=\frac{1}{2}sin2x$.

We now have the following integral:

$\int_{0}^{\frac{\pi}{2}}{\left(xsinx\right)}\,dx+\ frac{1}{2}\int_{0}^{\frac{\pi}{2}}sin2x\,dx$.

You will need to apply integration by parts to the first integral, and directly integrate the second.

Integration by parts : $\int u\,dv=uv-\int v\,du$.

let $u=x$ and $dv=sinx\,dx$.
$\therefore du=dx$ and $v=-cosx$.

Now we have:

$\int xsinx\,dx=-xcosx+\int cosx\,dx$
$\therefore \int xsinx\,dx=-xcosx+sinx+C$.

Since we're dealing with definite integrals, we don't need the $C$.

Now, we have the following:

$\int_{0}^{\frac{\pi}{2}}{\left(xsinx\right)}\,dx+\ frac{1}{2}\int_{0}^{\frac{\pi}{2}}sin2x\,dx={\left (-xcosx+sinx-\frac{1}{4}cos2x\right)}_{0}^{\frac{\pi}{2}}$.

Evaluating, we get:

${\left(-{\left(\frac{\pi}{2}\right)}cos{\left(\frac{\pi}{2 }\right)}+sin{\left(\frac{\pi}{2}\right)}-\frac{1}{4}cos2{\left(\frac{\pi}{2}\right)}\right) }-{\left(-{\left(0\right)}cos{\left(0\right)}+sin{\left(0\ri ght)}-\frac{1}{4}cos2{\left(0\right)}\right)}$
$={\left(1+\frac{1}{4}\right)}-{\left(-\frac{1}{4}\right)}=\frac{3}{2}$.

Hope this makes sense! If you want me to clarify, let me know!