Exercise 7.3 answers: Integrals

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1. Use sin²θ = (1 − cos 2θ)/2 with θ = 2x + 5, so 2θ = 4x + 10.
2. sin²(2x + 5) = (1 − cos(4x + 10))/2.
3. Integrate term by term: (1/2)∫dx − (1/2)∫cos(4x + 10) dx.
4. (1/2)∫dx = x/2, and ∫cos(4x + 10) dx = (1/4) sin(4x + 10).

Answerx/2 − (1/8) sin(4x + 10) + C

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1. Use sin A cos B = (1/2)[sin(A + B) + sin(A − B)] with A = 3x, B = 4x.
2. sin 3x cos 4x = (1/2)[sin 7x + sin(−x)] = (1/2)[sin 7x − sin x].
3. Integrate term by term: (1/2)∫sin 7x dx − (1/2)∫sin x dx.
4. ∫sin 7x dx = −(1/7) cos 7x, and ∫sin x dx = −cos x.

Answer−(1/14) cos 7x + (1/2) cos x + C

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1. Combine the last two factors: cos 4x × cos 6x = ½(cos 2x + cos 10x), using cos A cos B = ½[cos(A−B) + cos(A+B)].
2. The integrand becomes cos 2x × ½(cos 2x + cos 10x) = ½cos²2x + ½ cos 2x cos 10x.
3. Write cos²2x = ½(1 + cos 4x).
4. Write cos 2x cos 10x = ½(cos 8x + cos 12x), the same product rule again.
5. Put it together: the integrand is ¼(1 + cos 4x + cos 8x + cos 12x).
6. Integrate term by term, using ∫cos kx dx = (sin kx)/k.

Answerx/4 + (sin 4x)/16 + (sin 8x)/32 + (sin 12x)/48 + C

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1. This is an odd power of sine, so peel off one factor: sin³u = sin u × sin²u, where u = 2x + 1.
2. Replace sin²u using sin²u = 1 − cos²u, so sin³u = sin u − sin u cos²u.
3. Both pieces now integrate directly: ∫sin u du = −cos u, and ∫sin u cos²u du = −cos³u/3 (since the derivative of cos u is −sin u).
4. So in u, the integral is −cos u + cos³u/3.
5. Since u = 2x + 1 has rate 2, divide by 2: −cos u/2 + cos³u/6.

Answer−cos(2x+1)/2 + cos³(2x+1)/6 + C

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1. Group sin x cos x together: sin³x cos³x = (sin x cos x)³.
2. Use sin x cos x = (sin 2x)/2, so (sin x cos x)³ = sin³(2x)/8.
3. This is an odd power of sine again: sin³(2x) = sin 2x − sin 2x cos²2x.
4. Integrate each piece in the new variable u = 2x: ∫sin u du = −cos u, ∫sin u cos²u du = −cos³u/3, then divide by the rate 2.
5. So ∫sin³2x dx = −cos 2x/2 + cos³2x/6, and the whole integral is one eighth of that.

Answer−cos 2x/16 + cos³2x/48 + C

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1. Combine the first and third factors: sin x sin 3x = ½(cos 2x − cos 4x), using sin A sin B = ½[cos(A−B) − cos(A+B)].
2. Multiply by the middle factor: sin 2x × ½(cos 2x − cos 4x) = ½ sin 2x cos 2x − ½ sin 2x cos 4x.
3. Use sin 2x cos 2x = (sin 4x)/2.
4. Use sin 2x cos 4x = ½(sin 6x + sin(−2x)) = ½(sin 6x − sin 2x), since sine is an odd function.
5. Add it up: the integrand is ¼ sin 4x − ¼ sin 6x + ¼ sin 2x.
6. Integrate term by term using ∫sin kx dx = −(cos kx)/k.

Answer−(cos 2x)/8 − (cos 4x)/16 + (cos 6x)/24 + C

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1. The two sines have different arguments (4x and 8x), so use sin A sin B = ½[cos(A−B) − cos(A+B)].
2. Here A = 4x, B = 8x, so A − B = −4x and A + B = 12x.
3. sin 4x sin 8x = ½[cos(−4x) − cos 12x] = ½[cos 4x − cos 12x], since cosine is an even function.
4. Integrate each cosine using ∫cos kx dx = (sin kx)/k.

Answer(sin 4x)/8 − (sin 12x)/24 + C

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1. Use the half-angle forms 1 − cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2).
2. The fraction becomes sin²(x/2)/cos²(x/2) = tan²(x/2).
3. Write tan²(x/2) = sec²(x/2) − 1.
4. ∫sec²(x/2) dx = 2 tan(x/2), and ∫1 dx = x.

Answer2 tan(x/2) − x + C

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1. Write cos x = (1 + cos x) − 1, so the fraction splits as 1 − 1/(1 + cos x).
2. Use 1 + cos x = 2cos²(x/2), so 1/(1 + cos x) = ½ sec²(x/2).
3. So the integrand is 1 − ½ sec²(x/2).
4. ∫1 dx = x, and ∫½ sec²(x/2) dx = tan(x/2).

Answerx − tan(x/2) + C

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1. Write sin⁴x = (sin²x)².
2. Use sin²x = (1 − cos 2x)/2, so sin⁴x = ¼(1 − cos 2x)².
3. Expand: (1 − cos 2x)² = 1 − 2cos 2x + cos²2x.
4. Drop the power again: cos²2x = (1 + cos 4x)/2.
5. Add everything up: sin⁴x = ¼[1 − 2cos 2x + ½ + ½cos 4x] = 3/8 − ½cos 2x + ⅛cos 4x.
6. Integrate term by term.

Answer3x/8 − (sin 2x)/4 + (sin 4x)/32 + C

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1. Write cos⁴2x = (cos²2x)², with u = 2x.
2. Use cos²u = (1 + cos 2u)/2, so cos²2x = (1 + cos 4x)/2.
3. Square it: cos⁴2x = ¼(1 + cos 4x)² = ¼[1 + 2cos 4x + cos²4x].
4. Drop the power again: cos²4x = (1 + cos 8x)/2.
5. Add everything up: cos⁴2x = 3/8 + ½cos 4x + ⅛cos 8x.
6. Integrate term by term.

Answer3x/8 + (sin 4x)/8 + (sin 8x)/64 + C

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1. Write sin²x = 1 − cos²x = (1 − cos x)(1 + cos x).
2. The (1 + cos x) in the numerator and denominator cancel, leaving 1 − cos x.
3. ∫1 dx = x, and ∫cos x dx = sin x.

Answerx − sin x + C

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1. Use the double angle formula: cos 2x − cos 2α = 2cos² x − 1 − (2cos² α − 1) = 2(cos² x − cos² α)
2. Factor: 2(cos² x − cos² α) = 2(cos x − cos α)(cos x + cos α)
3. Simplify the fraction: (cos 2x − cos 2α) / (cos x − cos α) = 2(cos x + cos α)
4. Integrate: ∫ 2(cos x + cos α) dx = 2sin x + 2x cos α

Answer2sin x + 2x cos α + C

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1. Use 1 + sin 2x = sin²x + cos²x + 2 sin x cos x = (sin x + cos x)².
2. So the integrand is (cos x − sin x)/(sin x + cos x)².
3. Let u = sin x + cos x. Its rate of change is cos x − sin x — exactly the numerator.
4. So the integral becomes ∫du/u², which is −1/u.

Answer−1/(sin x + cos x) + C

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1. Let u = 2x. Write tan³u sec u = tan²u × (sec u tan u).
2. Use tan²u = sec²u − 1, so the integrand (in u) is (sec²u − 1)(sec u tan u).
3. Since the rate of change of sec u is sec u tan u, substitute t = sec u.
4. This becomes ∫(t² − 1) dt = t³/3 − t.
5. So in u: sec³u/3 − sec u. Since u = 2x has rate 2, divide by 2: sec³u/6 − sec u/2.

Answersec³(2x)/6 − sec(2x)/2 + C

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1. Write tan⁴x = tan²x × tan²x = tan²x(sec²x − 1).
2. Expand: tan⁴x = tan²x sec²x − tan²x.
3. Use tan²x = sec²x − 1 again in the second term: tan⁴x = tan²x sec²x − sec²x + 1.
4. The rate of change of tan x is sec²x, so ∫tan²x sec²x dx = tan³x/3.
5. ∫sec²x dx = tan x, and ∫1 dx = x.

Answertan³x/3 − tan x + x + C

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1. Split the fraction into two, one for each term on top: sin³x/(sin²x cos²x) + cos³x/(sin²x cos²x).
2. The first piece simplifies to sin x/cos²x = sec x tan x.
3. The second piece simplifies to cos x/sin²x = cosec x cot x.
4. Both are standard: ∫sec x tan x dx = sec x, and ∫cosec x cot x dx = −cosec x.

Answersec x − cosec x + C

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1. Use cos 2x = 1 − 2sin²x.
2. Then cos 2x + 2sin²x = 1 − 2sin²x + 2sin²x = 1.
3. So the whole fraction is simply 1/cos²x = sec²x.
4. ∫sec²x dx = tan x.

Answertan x + C

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1. Multiply top and bottom by sec⁴x: 1/(sin x cos³x) = sec⁴x/(tan x), since sin x cos³x × sec⁴x = tan x.
2. Write sec⁴x = sec²x × sec²x = (1 + tan²x) sec²x.
3. So the integrand is (1 + tan²x) sec²x/tan x.
4. Let t = tan x. Since its rate of change is sec²x, this becomes ∫(1 + t²)/t dt = ∫(1/t + t) dt.
5. That integrates to ln|t| + t²/2.

Answerln|tan x| + tan²x/2 + C

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1. Use cos 2x = cos²x − sin²x = (cos x − sin x)(cos x + sin x).
2. One factor of (cos x + sin x) cancels with the denominator, leaving (cos x − sin x)/(cos x + sin x).
3. Let u = sin x + cos x. Its rate of change is cos x − sin x — exactly the numerator.
4. So the integral becomes ∫du/u = ln|u|.

Answerln|sin x + cos x| + C

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1. Take x between 0 and π, so that π/2 − x lies between −π/2 and π/2 — the range sin⁻¹ is allowed to give back.
2. On this stretch, cos x = sin(π/2 − x), so sin⁻¹(cos x) = π/2 − x.
3. Integrate the straight line π/2 − x: ∫(π/2) dx − ∫x dx = (π/2)x − x²/2.

Answer(π/2)x − x²/2 + C

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1. a and b are fixed constants; only x changes.
2. Write 1 as sin[(x−a) − (x−b)]/sin(b−a) = [sin(x−a)cos(x−b) − cos(x−a)sin(x−b)]/sin(b−a).
3. Divide by cos(x−a)cos(x−b): the integrand becomes [tan(x−a) − tan(x−b)]/sin(b−a).
4. ∫tan(x−a) dx = −ln|cos(x−a)|, and ∫tan(x−b) dx = −ln|cos(x−b)|.
5. So the integral is [−ln|cos(x−a)| + ln|cos(x−b)|]/sin(b−a) = ln|cos(x−b)/cos(x−a)| / sin(b−a).

Answer[1/sin(b−a)] × ln|cos(x−b)/cos(x−a)| + C

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1. Split the fraction: sin²x/(sin²x cos²x) − cos²x/(sin²x cos²x).
2. That is 1/cos²x − 1/sin²x = sec²x − cosec²x.
3. ∫sec²x dx = tan x, and ∫cosec²x dx = −cot x, so ∫(−cosec²x) dx = cot x.
4. Adding: tan x + cot x.

Answer(A) tan x + cot x + C

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1. Question: ∫ eˣ(1 + x)/cos²(eˣx) dx.
2. Look at x·eˣ. Its derivative is eˣ + x·eˣ, which is eˣ(1 + x) — exactly the numerator.
3. So substitute u = x·eˣ. Then du = eˣ(1 + x) dx, and the integral becomes ∫ du/cos²u = ∫ sec²u du.
4. ∫ sec²u du = tan u + C.
5. Put u back as x·eˣ: the integral is tan(xeˣ) + C.
6. That is option (B).

Answer(B) tan (xeˣ) + C

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