More Integration Procedures, 5

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate

Solution:

The first thing that we have to do is to find the differential of the given equation above. If

then

Hence, the above equation becomes

but

Substitute the value of sec² x to the above equation, we have

Therefore,

Partial Fractions

Category: Algebra

"Published in Newark, California, USA"

Resolve into partial fractions for

Solution:

In this lesson, we need to learn this one because you will use this method often when you will study or take-up Integral Calculus. If you will integrate the algebraic fractions, then you must rewrite it into partial fractions first. If you know how to combine a fraction either by addition, subtraction, multiplication, or division, then you must know how to split a fraction into partial fractions. Anyway, let's consider the given equation above

The first thing that we have to do is to factor the numerator and denominator if you can. You must simplify a fraction into  lowest term always. That's a rule in Mathematics. As you notice that one of the factor in the denominator which is x² + 3x + 3 cannot be factored, then we have to leave it as is. In this case, we have to split the above equation into partial fractions as follows

Multiply both sides of the equation by their Least Common Denominator (LCD) which is (2x + 3)(x² + 3x + 3) as follows

Expand the right side of the equation and group according to their variables 

In order to solve for the value of A, B, and C, we need to equate the both sides of the equation according to their variables.

For x²: 0 = A + 2B
           A = - 2B                         (equation 1)

For x: 12 = 3A + 3B + 2C
          12 = 3(- 2B) + 3B + 2C
          12 = - 6B + 3B + 2C
          12 = - 3B + 2C                (equation 2)

For x⁰: 21 = 3A + 3C
             7 = A + C
             7 = - 2B + C                 (equation 3)         

Use equation 2 and equation 3 to solve for the value of B as follows

      - 3B + 2C = 12     —————> - 3B + 2C = 12                  
    -2(- 2B + C = 7)                             4B - 2C = - 14
                                                    ————————
                                                                  B = - 2

Substitute the value of B to equation 3, we have

         - 2B + C = 7
     - 2(- 2) + C = 7
              4 + C = 7
                    C = 3

Substitute the value of B to equation 1, we have

                    A = - 2B
                    A = - 2(- 2)
                    A = 4

Therefore,

Solving Radical Equations, 2

Category: Algebra

"Published in Newark, California, USA"

Find the roots of the equation for

Solution:

In solving the radical equations, you have to be very careful and write down the equations very well because the roots of the radical equations can be real, extraneous, or both. Anyway, consider the given equation above

Transpose either one of the term to the right side of the equation

Divide both sides of the equation by their common factor which is (2x - 1) as follows

Square the both sides of the equation to eliminate the radical signs

Expand the equation above and solve for the value of x as follows

Next, substitute x = 0 to the given equation, we have

Since the root is not satisfied to the given equation because  both sides of the equation are not equal, then the root of the equation is extraneous. NO ANSWER for this problem.