MHB Integration by Parts for Cosine Squared: Is My Approach Correct?

[jamescv31]

Greetings :)

Well I wanted to seek help if my solution is on the right path, given as follows:

1) $$ \int cos ^2x dx $$

So my solution follows like this:

[math]
u = cos^2x [/math]
[math]du = 1/2 (1+cos(2x)) [/math]
[math]v = x[/math]
[math]dv = dx$$

but I've stuck when its in the $$u.v - \int v.du$$

$$cos^2 (x) - \int (x) (1+cos(2x)/2)$$

Is this a correct path?

 

[evinda]

(Wave)It is known that :

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

You can calculate easier the integral using the above equality.

$$\int \cos^2 x dx=\int \frac{1+ \cos{(2x)}}{2}dx =\int \frac{1}{2}dx+\int \frac{\cos(2x)}{2}dx= \frac{1}{2}x+ \frac{\sin(2x)}{4}+C$$

 

[topsquark]

1) $$ \int cos ^2x dx $$

So my solution follows like this:

[math]
u = cos^2x [/math]
[math]du = 1/2 (1+cos(2x)) [/math]
[math]v = x[/math]
[math]dv = dx[/math]

You have handled your u incorrectly.

If [math]u = cos^2(x)[/math] then [math]du = 2 \cdot cos(x) \cdot -sin(x)~dx[/math]

So your integration line reads:
[math]\int u~dv = u \cdot v - \int v~du[/math]

[math]\int cos^2(x)~dx = cos^2(x) \cdot x - \int x \cdot -2~sin(x)~cos(x)~dx[/math]

 

[Prove It]

Greetings :)

Well I wanted to seek help if my solution is on the right path, given as follows:

1) $$ \int cos ^2x dx $$

So my solution follows like this:

[math]
u = cos^2x [/math]
[math]du = 1/2 (1+cos(2x)) [/math]
[math]v = x[/math]
[math]dv = dx$$

but I've stuck when its in the $$u.v - \int v.du$$

$$cos^2 (x) - \int (x) (1+cos(2x)/2)$$

Is this a correct path?

If you really want to use integration by parts, you really should use $\displaystyle \begin{align*} u = \cos{(x)} \implies \mathrm{d}u = -\sin{(x)}\,\mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \cos{(x)}\,\mathrm{d}x \implies v = \sin{(x)} \end{align*}$ to get

$\displaystyle \begin{align*} \int{\cos^2{(x)}\,\mathrm{d}x} &= \sin{(x)}\cos{(x)} - \int{ -\sin^2{(x)}\,\mathrm{d}x } \\ \int{\cos^2{(x)}\,\mathrm{d}x} &= \sin{(x)}\cos{(x)} + \int{\sin^2{(x)}\,\mathrm{d}x} \\ \int{\cos^2{(x)}\,\mathrm{d}x} &= \sin{(x)}\cos{(x)} + \int{ 1 - \cos^2{(x)}\,\mathrm{d}x} \\ \int{\cos^2{(x)}\,\mathrm{d}x} &= \sin{(x)}\cos{(x)} + \int{1\,\mathrm{d}x} - \int{\cos^2{(x)}\,\mathrm{d}x} \\ 2\int{\cos^2{(x)}\,\mathrm{d}x} &= \sin{(x)}\cos{(x)} + x \\ \int{\cos^2{(x)}\,\mathrm{d}x} &= \frac{1}{2}\sin{(x)}\cos{(x)} + \frac{1}{2}x \end{align*}$

with an integration constant on the end of course...

As you can see, this was the long way about it though, using the double angle identity for cosine would have made this a much easier problem...