Key Idea :
Let us consider the integration of a function with respect to x. If the derivative of a part of the function is at somewhere of the function and you are able to multiply the derivative along with dx using multiplication, then you can integration the function using substitution.
Consider the following integration.
In the function, the power of 'e' is x^{3}. The derivative is x^{3} is 3x^{2} and it is being as a part of the function and we will be able to write the 3x^{2} along with dx using multiplication.
Now substitute a new variable for x^{3} for which we have the derivative 3x^{2} being a part of the given function.
Let u = x^{3}.
u = x^{3}
Differentiate with respect to x on both sides.
du/dx = 3x^{2}
Multiply both sides by dx.
du = 3x^{2}dx
In the given integration, substitute x^{3} = u and 3x^{2}dx = du.
= ∫e^{u}du
= e^{u} + C
Substitute u = x^{3}.
Integrate each of the following with respect to x :
Example 1 :
x/√1 + x^{2}
Solution :
= ∫(x/√1 + x^{2})dx
= ∫(xdx)/√1 + x^{2} ----(1)
Let t = 1 + x^{2}.
t = 1 + x^{2}
Differentiate with respect to x.
dt/dx = 0 + 2x
dt/dx = 2x
Multiply both sides by dx.
dt = 2xdx
Divide both sides by 2.
dt/2 = xdx
Substitute 1 + x^{2} = t and xdx = dt/2 in (1).
∫(xdx)/√1 + x^{2} = ∫(dt/2)/√t
= (1/2)∫dt/√t
= (1/2)∫dt/t^{-1/2}
= (1/2)∫t^{-1/2}dt
= (1/2)t^{-1/2 + 1}/(-1/2 + 1) + C
= (1/2)t^{1/2}/(1/2) + C
= (1/2)t^{1/2}(2/1) + C
= t^{1/2} + C
= √t + C
Substitute t = x^{2} + 1.
= √(1 + x^{2}) + C
Example 2 :
x^{2}/(1 + x^{6})
Solution :
= ∫[x^{2}/(1 + x^{6})]^{ }dx
= ∫(x^{2}dx)/[1 + (x^{3})^{2}] ----(1)
Let t = x^{3}.
t = x^{3}
Differentiate with respect to x.
dt/dx = 3x^{2}
Multiply both sides by dx.
dt = 3x^{2}dx
Divide both sides by 3.
dt/3 = x^{2}dx
Substitute x^{3} = t and x^{2}dx = dt/3 in (1)
∫(x^{2}dx)/(1 + (x^{3})^{2} = ∫(dt/3)/(1 + t^{2})
= (1/3)∫1/(1 + t^{2})dt
= (1/3)tan^{-1}t + c
Substitute t = x^{3}.
= (1/3)tan^{-1}(x^{3}) + c
Example 3 :
(e^{x} - e^{-x})/(e^{x} + e^{-x})
Solution :
= ∫[(e^{x} - e^{-x})/(e^{x} + e^{-x})]dx
= ∫(e^{x} - e^{-x})dx/(e^{x} + e^{-x}) ----(1)
Let t = e^{x} + e^{-x}.
t = e^{x} + e^{-x}
Differentiate with respect to x.
dt/dx = e^{x} + e^{-x}(-1)
dt/dx = e^{x} - e^{-x}
Multiply both sides by dx.
dt = (e^{x} - e^{-x})dx
Substitute e^{x} + e^{-x }= t and (e^{x} - e^{-x})dx = dt in (1).
∫(e^{x} - e^{-x})dx/(e^{x} + e^{-x}) = ∫dt/t
= ∫(1/t)dt
= lnt
Substitute t = e^{x} + e^{-x}.
= ln(e^{x} + e^{-x}) + c
Example 4 :
(10x^{9} + 10^{x}ln10)/(10^{x} + x^{10})
Solution :
= ∫[(10x^{9} + 10^{x}ln10)/(10^{x} + x^{10})]dx
= ∫(10x^{9} + 10^{x}ln10)dx/(10^{x} + x^{10}) ----(1)
Let u = 10^{x} + x^{10}.
u = 10^{x} + x^{10}
Differentiate with respect to x.
du/dx = 10^{x}ln10 + 10x^{9}
Multiply both sides by dx.
du = (10^{x}ln10 + 10x^{9})dx
du = (10x^{9 }+ 10^{x}ln10)dx
Substitute 10^{x} + x^{10 }= u and (10x^{9 }+ 10^{x}ln10)dx in (1).
= ∫du/u
= ∫(1/u)du
= lnu + c
Substitute u = 10^{x} + x^{10}.
= ln(10^{x} + x^{10}) + c
Example 5 :
(sin√x)/√x
Solution :
= ∫[(sin√x)/√x]dx
= ∫(sin√x)(dx/√x) ----(1)
Let u = √x.
u = √x
u = x^{1/2}
du/dx = (1/2)x^{1/2 - 1}
du/dx = (1/2)x^{-1/2}
du/dx = 1/(2x^{1/2})
du/dx = 1/(2√x)
Multiply both sides by 2dx.
2du = dx/√x
Substitute √x = u and dx/√x = 2du in (1).
∫(sin√x)(dx/√x) = ∫(sinu)(2du)
= 2∫sinudu
= 2(-cosu) + c
= -2cosu + c
Substitute u = √x.
= -2cos√x + c
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