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Harder Substitution Simplified Revision Notes

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8.2.6 Harder Substitution

When we talk about "harder substitution" in A Level Maths, we're usually referring to substitution in integration problems where the function to be integrated is more complex, and the substitution is less straightforward. Let's walk through an example to illustrate the concept.


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📑Example

Problem

Integrate the following function with respect to xx :

x1+x2dx \int x \sqrt{1 + x^2} \, dx

Step-by-Step Solution


  1. Identify a suitable substitution: We notice that the expression 1+x21 + x^2 is inside the square root, and its derivative 2x2x is almost present in the integrand (we have xx ). This suggests that a good substitution could be:

u=1+x2u = 1 + x^2


  1. Differentiate  u\ u with respect to  x:\ x :

dudx=2x\frac{du}{dx} = 2x

Therefore, we can express dxdx in terms of dudu :

dx=du2xdx = \frac{du}{2x}


  1. Rewrite the integral in terms of uu : Substitute u=1+x2u = 1 + x^2 and dx=du2xdx = \frac{du}{2x} into the original integral:

x1+x2dx=xudu2x\int x \sqrt{1 + x^2} \, dx = \int x \sqrt{u} \cdot \frac{du}{2x}

Notice that the x x terms cancel out:

xudu2x=12udu\int x \sqrt{u} \cdot \frac{du}{2x} = \frac{1}{2} \int \sqrt{u} \, du

  1. Integrate with respect to uu : The integral now simplifies to:

12u12du\frac{1}{2} \int u^{\frac{1}{2}} \, du

Use the power rule for integration:

12u3232=1223u32=13u32\frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{1}{3} u^{\frac{3}{2}}


  1. Substitute back u=1+x2u = 1 + x^2 : Finally, replace uu with the original expression in terms of xx :

13(1+x2)32+C\frac{1}{3} (1 + x^2)^{\frac{3}{2}} + C

Where CC is the constant of integration.

Final Answer

x1+x2dx=13(1+x2)32+C\int x \sqrt{1 + x^2} \, dx = \frac{1}{3} (1 + x^2)^{\frac{3}{2}} + C

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If you haven't been doing so already, make sure you attempt the example questions!

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Example 1: Involving Trigonometric Functions

Question:

Evaluate sin(2x)1+cos(2x)dx\int \frac{\sin(2x)}{1 + \cos(2x)} \, dx .

Solution:

  1. Recognize the structure of the integrand: The expression 1+cos(2x) 1 + \cos(2x) suggests the use of a trigonometric identity. Use the identity:
cos(2x)=2cos2(x)1\cos(2x) = 2\cos^2(x) - 1

But here, using substitution directly is more straightforward.

  1. Let u=1+cos(2x)u = 1 + \cos(2x): This gives:
dudx=2sin(2x)ordu=2sin(2x)dx\frac{du}{dx} = -2\sin(2x) \quad \text{or} \quad du = -2\sin(2x) \, dx

We need sin(2x)dx\sin(2x) \, dx , so divide by 2-2:

du2=sin(2x)dx\frac{du}{-2} = \sin(2x) \, dx
  1. Rewrite the integral in terms of uu:
sin(2x)1+cos(2x)dx=du2u\int \frac{\sin(2x)}{1 + \cos(2x)} \, dx = \int \frac{\frac{du}{-2}}{u}
  1. Simplify and integrate:
du2u=12duu\int \frac{\frac{du}{-2}}{u} = -\frac{1}{2} \int \frac{du}{u}=12lnu+C= -\frac{1}{2} \ln |u| + C
  1. Substitute back u=1+cos(2x)u = 1 + \cos(2x):
12ln1+cos(2x)+C-\frac{1}{2} \ln |1 + \cos(2x)| + C

Thus, the solution is:

12ln1+cos(2x)+C-\frac{1}{2} \ln |1 + \cos(2x)| + C

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Example 2: Rational Functions

Question:

Evaluate 1x2x21dx\int \frac{1}{x^2 \sqrt{x^2 - 1}} \, dx.

Solution:

  1. Recognize the structure of the integrand: The term x21\sqrt{x^2 - 1} suggests a trigonometric substitution. We know that sec2(θ)1=tan2(θ)\sec^2(\theta) - 1 = \tan^2(\theta) .

  2. Substitute x=sec(θ)x = \sec(\theta): This gives:

dxdθ=sec(θ)tan(θ)sodx=sec(θ)tan(θ)dθ\frac{dx}{d\theta} = \sec(\theta)\tan(\theta) \quad \text{so} \quad dx = \sec(\theta)\tan(\theta) \, d\theta
  1. Rewrite the integral:
x2=sec2(θ)x^2 = \sec^2(\theta) x21=sec2(θ)1=tan(θ)\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta)

Substituting into the integral:

1sec2(θ)tan(θ)sec(θ)tan(θ)dθ\int \frac{1}{\sec^2(\theta) \cdot \tan(\theta)} \cdot \sec(\theta)\tan(\theta) \, d\theta
  1. Simplify:
1sec(θ)dθ=cos(θ)dθ\int \frac{1}{\sec(\theta)} \, d\theta = \int \cos(\theta) \, d\theta
  1. Integrate:
sin(θ)+C\sin(\theta) + C
  1. Substitute back θ=sec1(x)\theta = \sec^{-1}(x) : Since sin(θ)=x21x\sin(\theta) = \frac{\sqrt{x^2 - 1}}{x} , the final solution is:
x21x+C\frac{\sqrt{x^2 - 1}}{x} + C

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Example 3: Exponential and Polynomial Mix

Question:

Evaluate xex2dx\int x e^{x^2} \, dx.

Solution:

  1. Recognize the structure of the integrand: The term x2x^2 is in the exponent, and its derivative is 2x2x, which suggests substitution.

  2. Let u=x2u = x^2: This gives:

dudx=2xordu=2xdx\frac{du}{dx} = 2x \quad \text{or} \quad du = 2x \, dx

We need xdxx \, dx, so divide both sides by 2:

du2=xdx\frac{du}{2} = x \, dx
  1. Rewrite the integral:
xex2dx=eudu2\int x e^{x^2} \, dx = \int e^u \cdot \frac{du}{2}=12eudu= \frac{1}{2} \int e^u \, du
  1. Integrate:
12eu+C\frac{1}{2} e^u + C
  1. Substitute back u=x2u = x^2:
12ex2+C\frac{1}{2} e^{x^2} + C

Thus, the solution is:

12ex2+C\frac{1}{2} e^{x^2} + C

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Example 4: Involving Logarithms

Question:

Evaluate ln(x)xdx\int \frac{\ln(x)}{x} \, dx.

Solution:

  1. Recognize the structure of the integrand: The function ln(x)\ln(x) appears in the numerator, and its derivative is 1x\frac{1}{x}, suggesting substitution.

  2. Let u=ln(x)u = \ln(x): This gives:

dudx=1xordu=1xdx\frac{du}{dx} = \frac{1}{x} \quad \text{or} \quad du = \frac{1}{x} \, dx
  1. Rewrite the integral:
ln(x)xdx=udu\int \frac{\ln(x)}{x} \, dx = \int u \, du
  1. Integrate:
u22+C\frac{u^2}{2} + C
  1. Substitute back u=ln(x)u = \ln(x):
(ln(x))22+C\frac{(\ln(x))^2}{2} + C

Thus, the solution is:

(ln(x))22+C\frac{(\ln(x))^2}{2} + C

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Example 5: Multiple Substitutions

Question:

Evaluate dx(1+x2)3/2\int \frac{dx}{(1 + x^2)^{3/2}}.

Solution:

  1. Recognize the structure of the integrand: The term 1+x21 + x^2 suggests a trigonometric substitution since sec2(θ)\sec^2(\theta) or tan2(θ)\tan^2(\theta) often appear with such expressions.

  2. Substitute x=tan(θ)x = \tan(\theta): This gives:

dxdθ=sec2(θ)sodx=sec2(θ)dθ\frac{dx}{d\theta} = \sec^2(\theta) \quad \text{so} \quad dx = \sec^2(\theta) \, d\theta

Also, 1+x2=1+tan2(θ)=sec2(θ) 1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta).

  1. Rewrite the integral:
dx(1+x2)3/2=sec2(θ)dθ(sec2(θ))3/2=cos(θ)dθ\int \frac{dx}{(1 + x^2)^{3/2}} = \int \frac{\sec^2(\theta) \, d\theta}{(\sec^2(\theta))^{3/2}} = \int \cos(\theta) \, d\theta
  1. Integrate:
sin(θ)+C\sin(\theta) + C
  1. Substitute back θ=tan1(x)\theta = \tan^{-1}(x):
sin(θ)=x1+x2\sin(\theta) = \frac{x}{\sqrt{1 + x^2}}

Thus, the solution is:

x1+x2+C\frac{x}{\sqrt{1 + x^2}} + C

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