← Integration Techniques › Integration Using Trigonometric Identities › Solutions
Integration Using Trigonometric Identities — Full Worked Solutions
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Find ∫sin²(x) dx. Fluency
Identity used: sin²x = (1 − cos 2x)/2.
∫sin²x dx = ∫ (1 − cos 2x)/2 dx
= (1/2)∫1 dx − (1/2)∫cos 2x dx
= (1/2)x − (1/2) × (sin 2x)/2 + C
= x/2 − (sin 2x)/4 + C
Check: d/dx[x/2 − (sin 2x)/4] = 1/2 − (2 cos 2x)/4 = 1/2 − (cos 2x)/2 = (1 − cos 2x)/2 = sin²x ✓
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Find ∫cos²(x) dx. Fluency
Identity used: cos²x = (1 + cos 2x)/2.
∫cos²x dx = ∫ (1 + cos 2x)/2 dx
= (1/2)x + (1/2)(sin 2x)/2 + C
= x/2 + (sin 2x)/4 + C
Note: Compare with Q1 — the only difference is the sign of the (sin 2x)/4 term, reflecting the identity cos²x = 1 − sin²x. Adding Q1 and Q2 gives x + C, consistent with ∫(sin²x+cos²x)dx = ∫1 dx = x + C ✓
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Find ∫sin²(x)cos²(x) dx. Fluency
Strategy: Use sin x cos x = (sin 2x)/2 first.
sin²x cos²x = (sin x cos x)² = [(sin 2x)/2]² = sin²(2x)/4.
Apply sin²(2x) = (1 − cos 4x)/2:
sin²(2x)/4 = (1 − cos 4x)/8.
∫sin²x cos²x dx = ∫ (1 − cos 4x)/8 dx
= x/8 − (sin 4x)/32 + C
= x/8 − (sin 4x)/32 + C
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Find ∫sin³(x) dx. Fluency
Strategy (odd power): Write sin³x = sin x × sin²x and substitute u = cos x.
sin³x = sin x(1 − cos²x).
Let u = cos x, du = −sin x dx ⇒ sin x dx = −du.
∫sin³x dx = ∫(1 − u²)(−du) = −∫(1 − u²) du = −u + u³/3 + C
Substitute back u = cos x:
= −cos x + cos³(x)/3 + C
Check: d/dx[−cos x + cos³x/3] = sin x + 3cos²x(−sin x)/3 = sin x − sin x cos²x = sin x(1−cos²x) = sin³x ✓
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Find ∫cos&sup4;(x) dx. Understanding
Step 1: Write cos4x = (cos²x)² and apply cos²x = (1+cos 2x)/2:
cos4x = [(1+cos 2x)/2]² = (1 + 2cos 2x + cos²2x)/4.
Step 2: Apply cos²(2x) = (1+cos 4x)/2:
cos4x = (1 + 2cos 2x + (1+cos 4x)/2)/4 = (2 + 4cos 2x + 1 + cos 4x)/8 = (3 + 4cos 2x + cos 4x)/8.
Step 3: Integrate term by term:
∫cos4x dx = (1/8)∫(3 + 4cos 2x + cos 4x) dx
= (1/8)[3x + 2 sin 2x + (sin 4x)/4] + C
= 3x/8 + (sin 2x)/4 + (sin 4x)/32 + C
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Evaluate ∫0π/4 tan²(x) dx. Understanding
Identity used: tan²x = sec²x − 1 (from 1 + tan²x = sec²x).
∫0π/4 tan²x dx = ∫0π/4(sec²x − 1) dx
= [tan x − x]0π/4
At x = π/4: tan(π/4) − π/4 = 1 − π/4.
At x = 0: tan(0) − 0 = 0.
= 1 − π/4 ≈ 0.215
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Find ∫sin(3x)cos(x) dx. Understanding
Product-to-sum formula: sin A cos B = (1/2)[sin(A+B) + sin(A−B)].
With A = 3x, B = x:
sin(3x)cos(x) = (1/2)[sin(4x) + sin(2x)].
∫sin(3x)cos(x) dx = (1/2)∫[sin 4x + sin 2x] dx
= (1/2)[−(cos 4x)/4 − (cos 2x)/2] + C
= −(cos 4x)/8 − (cos 2x)/4 + C
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Find ∫cos²(2x)sin²(2x) dx. Understanding
Strategy: Use sin(2x)cos(2x) = (sin 4x)/2.
cos²(2x)sin²(2x) = [sin(2x)cos(2x)]² = [(sin 4x)/2]² = sin²(4x)/4.
Apply sin²(4x) = (1 − cos 8x)/2:
cos²(2x)sin²(2x) = (1 − cos 8x)/8.
∫cos²(2x)sin²(2x) dx = ∫(1 − cos 8x)/8 dx
= x/8 − (sin 8x)/64 + C
= x/8 − (sin 8x)/64 + C
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Find the area enclosed between y = sin²(x) and y = cos²(x) over [0, π/2]. Problem Solving
Find intersection: sin²x = cos²x ⇒ cos 2x = 0 ⇒ 2x = π/2 ⇒ x = π/4 on [0, π/2].
Identify which is larger:
- On [0, π/4]: cos x > sin x, so cos²x > sin²x. Difference = cos²x − sin²x = cos 2x > 0.
- On [π/4, π/2]: sin²x > cos²x. Difference = sin²x − cos²x = −cos 2x > 0.
Compute area by symmetry:
Area = 2∫0π/4(cos²x − sin²x) dx = 2∫0π/4 cos 2x dx
= 2[(sin 2x)/2]0π/4 = [sin 2x]0π/4
= sin(π/2) − sin(0) = 1 − 0
Area = 1 square unit
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Show ∫0π sin²(nx) dx = π/2 for any positive integer n. Hence evaluate ∫0π cos²(nx) dx. Problem Solving
Part 1 — Proof:
Apply sin²(nx) = (1 − cos(2nx))/2:
∫0π sin²(nx) dx = ∫0π (1 − cos 2nx)/2 dx
= (1/2)[x − (sin 2nx)/(2n)]0π
At x = π: (1/2)[π − sin(2nπ)/(2n)].
Since n is a positive integer, 2nπ is a multiple of 2π, so sin(2nπ) = 0.
At x = 0: (1/2)[0 − 0] = 0.
Result: (1/2)(π − 0) = π/2. QED.
Part 2 — Evaluating ∫0π cos²(nx) dx:
From the Pythagorean identity: sin²(nx) + cos²(nx) = 1.
Integrating both sides over [0, π]:
∫0πsin²(nx) dx + ∫0πcos²(nx) dx = π
π/2 + ∫0πcos²(nx) dx = π
∫0π cos²(nx) dx = π/2
This result is fundamental in Fourier series: on [0,π], each “basis function” sin(nx) or cos(nx) has squared integral equal to π/2, which is used to compute Fourier coefficients.