## Minggu, Juli 22, 2012

### SOAL n PEMBAHASAN INTEGRAL_5

NO-1
$\dpi{80} \fn_jvn Jika \:\: \: a =sin^{2}15^{0}\: \: dan\: \: b =cos^{2}15^{0}\: \: maka\: \:\int_{a}^{b}(x-a)(x-b).dx=$
$\dpi{80} \fn_jvn A.-\frac{\sqrt{3}}{16}$ (*)
$\dpi{80} \fn_jvn B.-\frac{\sqrt{3}}{8}$
$\dpi{80} \fn_jvn C.-\frac{\sqrt{3}}{6}$
$\dpi{80} \fn_jvn D.-\frac{\sqrt{3}}{4}$
$\dpi{80} \fn_jvn E.-\frac{\sqrt{3}}{2}$
dengan bantuan WolframAlfa
akan lebih asyik....
karena kita bisa langsung
masukin hasilnya
Bentuk soal diatas akan lebih mudah kita kerjakan dengan memakai INTEGRAL PARTIAL.
yaitu memakai rumus atau pakai tabel TANZALIN.
$\dpi{100} \fn_jvn Rumus\: {\color{Magenta} integral\: partial}\: \int u.dv=u.v-\int v.du$
$\dpi{100} \fn_jvn u=(x-a)\Rightarrow du=dx$
$\dpi{100} \fn_jvn dv=(x-b).dx\Rightarrow v=\frac{1}{2}(x-b)^{2}$
$\dpi{100} \fn_jvn (x-a).(x-b)^{2}-\int \frac{1}{2}(x-b)^{2}.dx$
$\dpi{100} \fn_jvn (x-a).(x-b)^{2}- \frac{1}{2}.\frac{1}{3}(x-b)^{3}\mid _{a}^{b}$   $\dpi{100} \fn_jvn =\frac{1}{6}(a-b)^{3}$
$\dpi{100} \fn_jvn =\frac{1}{6}(sin^{2}15^{0}-cos^{2}15^{0})^{3}=-\frac{1}{6}.(cos^{2}15^{0}-sin^{2}15^{0})^{3}$
$\dpi{100} \fn_jvn -\frac{1}{6}.(cos\, 30^{0})^{3}=-\frac{1}{6}.(\frac{\sqrt{3}}{2})^{3}=-\frac{1}{6}.(\frac{3\sqrt{3}}{8})=-\frac{\sqrt{3}}{16}$

Beberapa contoh soal yang memakai teknik INTEGRAL PARTIAL
bisa di simak dibawah ini yang aq ambilkan di alamat http://www.intmath.com/blog/tanzalin-method-for-easier-integration-by-parts/4339

### Example 1

$\int2x(3x-2)^6dx$

#### Integration by Parts Method

First, let’s see normal Integration by Parts for comparison.
We identify u, v, du and dv as follows:
 u = 2x dv = (3x − 2)6dx du = 2dx $v=\frac{1}{21}(3x-2)^7$
Integration by Parts then gives us:
$\int{udv}=uv-\int{vdu}$
$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{2}{21}\int(3x-2)^7dx$
Now, we find the unknown integral:
$\int(3x-2)^7dx=\frac{1}{24}(3x-2)^8+C$
Putting it together, we have:
$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{1}{252}(3x-2)^8+C$
We can then factor and simplify this to give:
$\int2x(3x-2)^6dx=\frac{21x+2}{252}(3x-2)^7+C$
The Tanzalin Method is somewhat less messy.

#### Example 1, now using Tanzalin Method

In the Tanzalin Method, we set up a table as follows. In the first column are successive derivatives of the simplest polynomial term of our integral. (We need to choose this term for the derivatives column because it will disappear after a few steps.)
In the second column are the integrals of the second term of the integral.
We just multiply the 2 terms with green background in the table (the original 2x term and the first integral term). We don’t change the sign of this term.
We then multiply the 2 terms with yellow background (the first derivative and the second integral term). We assign a negative sign to the product, as shown.
The answer for the integral is just the sum of the 2 terms in the final column.
The question again, for reference:
$\int2x(3x-2)^6dx$
DerivativesIntegralsSignSame-color Products
2x(3x − 2)6
2$\frac{1}{21}(3x-2)^7$+$\frac{2x}{21}(3x-2)^7$
0$\frac{1}{504}(3x-2)^8$$-\frac{1}{252}(3x-2)^8$
Summing the 4th column:
$\int2x(3x-2)^6dx=\frac{2x}{21}(3x-2)^7-\frac{1}{252}(3x-2)^8+C$
(We add the constant of integration, C only at the end, not in the table.)
We can then factor and simplify this to give:
$\int2x(3x-2)^6dx=\frac{21x+2}{252}(3x-2)^7+C$

### Example 2

$\int{x}\sin{x}\hspace{3}dx$
We’ll go straight to the Tanzalin Method.
DerivativesIntegralsSignSame-color Products
xsin x
1−cos x+x cos x
0−sin xsin x
We multiplied (x) by (−cos x) and we didn’t change the sign.
We then multiplied (1) by (−sin x) and changed the sign.
$\int{x}\sin{x}\hspace{3}dx=-x\cos{x}+\sin{x}+C$

### Example 3

$\int{x^2}\sqrt{x-1}\hspace{3}dx$
Using the Tanzalin Method requires 4 rows in the table this time, since there is one more derivative to find in this case.
We need to alternate the signs (3rd column), so our 4th row will have a positive sign.
DerivativesIntegralsSignSame-color Products
x2$(x-1)^\frac{1}{2}$
2x$\frac{2}{3}(x-1)^\frac{3}{2}$+$\frac{2x^2}{3}(x-1)^\frac{3}{2}$
2${\frac{4}{15}(x-1)^{\frac{5}{2}}}$${-\frac{8x}{15}(x-1)^{\frac{5}{2}}}$
0${\frac{8}{105}(x-1)^{\frac{7}{2}}}$+${\frac{16}{105}(x-1)^{\frac{7}{2}}}$
${\int{x^2\sqrt{x-1}}dx=}{\frac{2x^2}{3}(x-1)^{\frac{3}{2}}-}{\frac{8x}{15}(x-1)^{\frac{5}{2}}+}{\frac{16}{105}(x-1)^{\frac{7}{2}}}+C$

### Example 4 – a Problem Arises

This is the same question as Example 3 in the Integration by Partssection in IntMath.
$\int{x^2}\ln{4x}\hspace{4}dx$
We need to choose ln 4x (natural logarithm of 4x) for the first column this time, following the Integration by Parts priority recommendations of:
1. log of x,
2. raised to power x
3. x raised to a power
[Note: If we choose the other way round, we would have to find integrals of ln 4x, which is not pretty (and certainly no easier than doing it all using Integration by Parts. See Example 6 on this page:Integration by Parts).
DerivativesIntegralsSignSame-color Products
ln 4xx2
$\frac{1}{x}$ $\frac{x^3}{3}$+$\frac{x^3\ln4x}{3}$
$-\frac{1}{x^2}$$\frac{x^4}{12}$$-\frac{x^3}{12}$
$\frac{2}{x^3}$$\frac{x^5}{60}$+$-\frac{x^3}{60}$
When do we stop? The derivatives column will continue to grow, as will the integrals column. The Tanzalin Method requires one of the columns to "disappear" (have value 0) so we have somewhere to stop.
$\int{x^2}\ln{4x}\hspace{4}dx=\frac{x^3\ln4x}{3}-\left(\frac{1}{12}+\frac{1}{60}+\frac{1}{180}+\frac{1}{420}+\ldots\right)x^3+C$
That expression in brackets must equal $\frac{1}{9}$ (since this is the answer we got using Integration by Parts), but as you can see, it is not a Geometric Progression and would take some figuring out.