Q.Evaluate ∫0πa2cos2x+b2sin2xxdx
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Definite Integral Symmetry: The Shortcut That Saves You Work
Asked to find the area under f(x)=x3 from x=−2 to x=2? You could integrate directly — but there's a much faster way if you notice the symmetry of the graph.
The Intuition: What Does "Symmetry" Mean Here?
A function can be symmetric about the y-axis (like x2 or cosx) or about the origin (like x3 or sinx). When you integrate over a symmetric interval — from −a to a — these symmetries create a predictable cancellation or doubling.
Even functions (symmetric about the y-axis): f(−x)=f(x) — think x2, x4, cosx, ∣x∣. The left side mirrors the right, so the area from −a to 0 equals the area from 0 to a. The total is double the area on one side.
Odd functions (symmetric about the origin): f(−x)=−f(x) — think x3, x5, sinx, tanx. The left side is the negative mirror of the right, so every positive area on the right is cancelled by an equal negative area on the left. The total is zero.
This only works when the limits are symmetric about zero — from −a to a. For [0,a] or [1,3], symmetry doesn't help directly.
The Precise Statement
Let f be continuous on [−a,a].
- If f is even (f(−x)=f(x)), then ∫−aaf(x)dx=2∫0af(x)dx
- If f is odd (f(−x)=−f(x)), then ∫−aaf(x)dx=0
∫−aaf(x)dx={2∫0af(x)dx0if f is evenif f is odd
Why Does This Work? (A Quick Proof)
Split at zero:
∫−aaf(x)dx=∫−a0f(x)dx+∫0af(x)dx
For the first term substitute u=−x (dx=−du; x=−a→u=a, x=0→u=0):
∫−a0f(x)dx=∫0af(−u)du
Now use symmetry:
- If f is even, f(−u)=f(u), so this becomes ∫0af(u)du; adding the second term gives 2∫0af(x)dx.
- If f is odd, f(−u)=−f(u), so this becomes −∫0af(u)du; adding the second term gives 0.
Common Mistakes to Avoid
Don't assume symmetry without checking. A "balanced"-looking function need not be even or odd — e.g. f(x)=x2+x is neither, so these formulas don't apply. Always verify f(−x)=f(x) or f(−x)=−f(x) for all x.
The interval must be [−a,a]. If the limits are [−2,3], symmetry doesn't apply directly — split the integral or shift variables.
Examples to Cement the Idea
Example 1: ∫−33x4dx — x4 is even, so
∫−33x4dx=2∫03x4dx=2[5x5]03=2⋅5243=5486 …
Use the King property ∫0af(x)dx=∫0af(a−x)dx to remove the x in the numerator.
Let I=∫0πa2cos2x+b2sin2xxdx. Replacing x→π−x leaves the denominator unchanged (since cos2(π−x)=cos2x, sin2(π−x)=sin2x):
I=∫0πa2cos2x+b2sin2xπ−xdx.
Adding the two forms:
2I=π∫0πa2cos2x+b2sin2xdx=πJ.
Evaluate J. By symmetry about x=2π, J=2∫0π/2a2cos2x+b2sin2xdx. Divide by cos2x and put t=tanx: …
The King property ∫0af(x)dx=∫0af(a−x)dx kills the x in the numerator, reducing the problem to a standard integral. The value is 2abπ2.
The idea
The denominator a2cos2x+b2sin2x is symmetric about x=2π, but the numerator is just x. The substitution x→π−x is designed exactly to exploit that mismatch.
Step 1 — Apply the King property
Let
I=∫0πa2cos2x+b2sin2xxdx.
Using ∫0af(x)dx=∫0af(a−x)dx with a=π, replace x by π−x. Since cos(π−x)=−cosx and sin(π−x)=sinx, the denominator is unchanged:
I=∫0πa2cos2x+b2sin2xπ−xdx.
Step 2 — Add the two forms
Adding the original and transformed integrals, the x and −x cancel:
2I=∫0πa2cos2x+b2sin2xx+(π−x)dx=π∫0πa2cos2x+b2sin2xdx.
Call the remaining integral J, so 2I=πJ.
Step 3 — Evaluate J
The integrand is symmetric about x=2π, so
J=2∫0π/2a2cos2x+b2sin2xdx.
Divide numerator and denominator by cos2x: …
Method: King's property to kill the x, then a tan substitution
Use this for ∫0π⋯x⋅(even-about-2π function)dx: the reflection property removes the linear x, reducing the problem to a pure trig-rational integral.
Steps
Step 1: Apply x→a−x (here a=π).
Because cos2(π−x)=cos2x and sin2(π−x)=sin2x, the denominator is unchanged and
I=∫0πa2cos2x+b2sin2xπ−xdx.
Step 2: Add the two forms to eliminate x.
x+(π−x)=π, so
2I=π∫0πa2cos2x+b2sin2xdx=πJ.
Step 3: Evaluate the remaining J by a tan substitution. …
Common Mistakes
Mistake 1: Thinking the substitution x→π−x changes the denominator.
Why it's wrong: although cos(π−x)=−cosx, the denominator uses cos2x, so squaring removes the sign and it is unchanged — that invariance is exactly what makes the trick work. Correct approach: note the denominator depends only on cos2x,sin2x, both even under x→π−x.
Mistake 2: Forgetting the 21 / limit handling when evaluating J. …
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