Q.By using the properties of definite integrals, evaluate the integral ∫02πcos5xdx
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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 symmetry over the full period. First, since cos5(2π−x)=cos5x,
∫02πcos5xdx=2∫0πcos5xdx.
Next, cos5(π−x)=(−cosx)5=−cos5x, so on [0,π] the function is odd about x=2π and …
Over a full period the positive and negative loops of an odd power of cosine cancel exactly, so ∫02πcos5xdx=0.
The idea
We use two definite-integral properties:
∫02af(x)dx=2∫0af(x)dxif f(2a−x)=f(x),
∫0af(x)dx=0if f(a−x)=−f(x).
1. Fold [0,2π] onto [0,π]
With a=π, check f(2π−x)=cos5(2π−x)=cos5x=f(x). So
∫02πcos5xdx=2∫0πcos5xdx.
2. Show the half-integral is zero …
Method: Fold a full-period integral, then use half-interval sign symmetry
For an odd power of cos (or sin) over a full period, combine two reflection properties: first fold [0,2a] onto [0,a], then show the half-integral vanishes by a sign flip.
Steps
Step 1: Fold using f(2a−x)=f(x).
If cosn(2π−x)=cosnx, then ∫02π=2∫0π.
Step 2: Test the half-interval for anti-symmetry.
Check f(a−x)=−f(x): since cos(π−x)=−cosx, an odd power gives cosn(π−x)=−cosnx. …
Common Mistakes
Mistake 1: Assuming an integral over a full period is automatically zero.
Why it's wrong: ∫02πcos2xdx=π=0 — only odd powers cancel; even powers have positive net area. Correct approach: it is the odd power (and the sign flip cos(π−x)=−cosx) that forces 0 here.
Mistake 2: Mishandling the folding property. …
Showing the 12 most recent of 15 on this concept.
- GUJCET 2024Set 131 markMCQQ.∫−π/2π/2(x5−x3cosx+sin3x−3)dx= __________. (A) 0 (B) 3π (C) −3π (D) −π
›Reveal solutionSolution
Odd functions vanish over [−a,a]; only the constant −3 contributes.
Concept. ∫−aa(odd)dx=0; ∫−aa(even)dx=2∫0a.
Over [−π/2,π/2]: x5 (odd), x3cosx (odd), sin3x (odd) all integrate to 0. …
- GUJCET 2019Set 171 markMCQQ.∫−π/2π/21+25xcos22xdx=. (A) −2π (B) 2π (C) 4π (D) −4π
›Reveal solutionSolution
The identity ∫−aa1+txf(x)dx=21∫−aaf(x)dx (for even f) removes the 25x factor.
Concept. For even f, ∫−aa1+txf(x)dx=21∫−aaf(x)dx. Here f(x)=cos22x is even.
Steps.
- I=21∫−π/2π/2cos22xdx=∫0π/2cos22xdx. …
- GUJCET 2022Set 081 markMCQQ.∫−π/2π/2(x13+xcosx+tan15x+1)dx= ______. (A) 1 (B) 2 (C) π (D) 0
›Reveal solutionSolution
x13, xcosx and tan15x are odd → integrate to 0 over a symmetric interval; the constant 1 gives π. …
- GUJCET 2021Set 151 markMCQQ.∫−11cot−1xdx= (A) 0 (B) 2π (C) π (D) 2π
›Reveal solutionSolution
Use cot−1(−x)=π−cot−1x with the symmetric-interval property.
Concept: …
- GUJCET 2020Set 071 markMCQQ.∫−2π2πlog(2019+x2019−x)dx= ________. (A) π (B) 0 (C) 2π (D) 1
›Reveal solutionSolution
log2019+x2019−x is odd, and ∫−aa(odd)=0.
Concept: Let g(x)=log2019+x2019−x. Then
g(−x)=log2019−x2019+x=−log2019+x2019−x=−g(x) …
- GUJCET 2022Set 081 markMCQQ.∫01tan−1(1+x−x22x−1)dx= ______. (A) 4π (B) 0 (C) −1 (D) 1
›Reveal solutionSolution
1+x−x22x−1=1+x(1−x)x−(1−x), so the integrand is tan−1x−tan−1(1−x).
Concept. Using tan−1A−tan−1B=tan−11+ABA−B with A=x, B=1−x, the integrand equals tan−1x−tan−1(1−x). …
- GUJCET 2023Set 091 markMCQQ.∫π/6π/31+tan4x1dx= ______. (A) 12π (B) 4π (C) 2π (D) 6π
›Reveal solutionSolution
Add the integral to its reflection using tan(π/2 − x) = cot x; the integrand pair sums to 1.
Concept. ∫abf(x)dx=∫abf(a+b−x)dx, with a+b=6π+3π=2π.
Solution. Let I=∫π/6π/31+tan4xdx. Replacing x→2π−x turns tan4x into cot4x: …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.∫−π/4π/4(x2sinx+tan3x−1)dx= ____.(a) 0(b) −π/2(c) π/2(d) π/4
›Reveal solutionSolution
The odd-function terms vanish on the symmetric interval [−π/4,π/4], leaving only the constant term.
x2sinx is odd (even × odd), and tan3x is odd, so both integrate to 0 over [−π/4,π/4].
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.∫−11sin7x⋅cos6xdx= ______.(a) −1(b) 2(c) 0(d) 1
›Reveal solutionSolution
The integrand is an odd function, and the interval is symmetric about 0.
sin7x is odd and cos6x is even, so sin7xcos6x is odd.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.∫02πsin3xcos2xdx= ______.(a) 2π(b) −1(c) 1(d) 0
›Reveal solutionSolution
Substitute u=cosx and note the limits give the same u-value.
sin3xcos2x=sinx(1−cos2x)cos2x. With u=cosx, du=−sinxdx:
∫sin3xcos2xdx=−∫(u2−u4)du=−3u3+5u5.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.The value of ∫−π/2π/2(x3+cosx+tan5x)dx is ___.(a) 0(b) 2(c) π(d) 1
›Reveal solutionSolution
Use odd/even symmetry over [−2π,2π].
Over a symmetric interval [−a,a], the integral of an odd function is 0 and of an even function is 2∫0a.
x3 is odd ⇒0; tan5x is odd ⇒0; cosx is even. …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.∫−11sin5xcos4xdx= ___.(a) 0(b) 2(c) −2(d) 3
›Reveal solutionSolution
The integrand is odd, so integrating over the symmetric interval gives zero.
…
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