Page 11 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes (x 4  5) 3/2
Thus is very big by a factor of 4. The exact anti derivative is
3/2
1 (x 4  5) 1/2 1 4 3/2
 (x  5)
4 3/2 6
Thus
1 4 3/2
3
4
 x x  5 dx  6 (x  5) –C
As a concluding check:

d  1 4 3/2 1 3 4 1/2 4 3 4 1/2
 (x  5)  – . (x  5) 4x – x (x  5)
dx  6  6 2

Notes As we have observed in the previous examples, anti differentiating a function
frequently includes “correcting for” constant factors: if differentiation generates an extra
1
factor of 4, anti differentiation will need a factor of .
4
1.2.2 The Method of Substitution

When the integrand is intricated, it assists to formalize this guess-and-check method as below:

To Make a Substitution

dw
Let w be the “inside function” and dw  w '( )dx  dx .
x
dx
Let’s do again the first example by means of a substitution.


Example: Find 3x 2 cos(x 3 )dx .
Solution:
As before, we gaze for an inside function whose derivative occurs—in this case x . Let w = x .
3
3
2
x
Then dw  w '( )dx  3x dx . The original integrand can now be entirely rewritten in terms of the
new variable w:
2
(x 3 ) 3x dx 3
2
3
w
 3x cos(x )dx  cos .   coswdw  sin  C  sin(x )–C

w dw
By altering the variable to w, we can shorten the integrand. We now have cos w, which can be
anti differentiated more simply. The concluding step, after anti differentiating, is to convert
back to the original variable, x.
1.2.3 Trigonometric Substitutions
As we know Substitutions permit us to solve complicated-looking integrals by translating them
into something more manageable. Now, we shall scrutinize a different type of substitution: a
trigonometric substitution, which will permit us to integrate more functions.

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