Integration techniques worksheet with multiple problems involving different calculus methods.

Since all options (ExplainAnswer, TeachTopic, GeneratePractice) are false, I will only provide the final answers to each integral listed in the worksheet.

Let me solve each part carefully and concisely:

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1. Evaluate

(a) ∫ x³ dx
= x⁴/4 + C

(b) ∫ 2 / ∛(x²) dx = ∫ 2 x⁻²⁄³ dx
= 2 · x¹⁄³ / (1/3) + C = 6 x¹⁄³ + C

(c) ∫ 2/x dx = 2 ln|x| + C

(d) ∫ e²ˣ dx
Let u = 2x ⇒ du = 2 dx ⇒ dx = du/2
∫ eᵘ · (du/2) = (1/2)eᵘ + C = (1/2)e²ˣ + C

(e) ∫ e^(3t/5) dt
Let u = 3t/5 ⇒ du = 3/5 dt ⇒ dt = 5/3 du
∫ eᵘ · (5/3) du = (5/3)eᵘ + C = (5/3)e^(3t/5) + C
*(Note: variable is t, not x — but as written, it's e^(3t/5) dt)*
Assuming typo and it's dx: ∫ e^(3x/5) dx = (5/3)e^(3x/5) + C

(f) ∫ e⁻ˣ dx = –e⁻ˣ + C

(g) ∫ cos(5x) dx = (1/5) sin(5x) + C

(h) ∫ cos(2x) dx = (1/2) sin(2x) + C

(i) ∫ cos(x/3) dx = 3 sin(x/3) + C

(j) ∫ sin(4x) dx = –(1/4) cos(4x) + C

(k) ∫ sin(–2x) dx = –∫ sin(2x) dx = –[–(1/2)cos(2x)] + C = (1/2)cos(2x) + C
*(since sin(–θ) = –sinθ)*

(l) ∫ 1/(4 + x²) dx = (1/2) arctan(x/2) + C
*(standard form: ∫ 1/(a² + x²) dx = (1/a) arctan(x/a))*

(m) ∫ 1/(16 + x²) dx = (1/4) arctan(x/4) + C

(n) ∫ 1/(3 + x²) dx = (1/√3) arctan(x/√3) + C

(o) ∫ 1/√(4 – x²) dx = arcsin(x/2) + C
*(standard: ∫ 1/√(a² – x²) dx = arcsin(x/a))*

(p) ∫ 1/√(9 – x²) dx = arcsin(x/3) + C

(q) ∫ 1/√(3 – x²) dx = arcsin(x/√3) + C

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2. Evaluate (u-substitution)

(a) ∫ x² sin(x³ + 7) dx
Let u = x³ + 7 ⇒ du = 3x² dx ⇒ x² dx = du/3
= ∫ sin(u) · (du/3) = –(1/3) cos(u) + C = –(1/3) cos(x³ + 7) + C

(b) ∫ (1 + sin x)³ cos x dx
Let u = 1 + sin x ⇒ du = cos x dx
= ∫ u³ du = u⁴/4 + C = (1 + sin x)⁴ / 4 + C

(c) ∫ sin⁴x cos x dx
Let u = sin x ⇒ du = cos x dx
= ∫ u⁴ du = u⁵/5 + C = sin⁵x / 5 + C

(d) ∫ sec²x tan⁴x dx
Let u = tan x ⇒ du = sec²x dx
= ∫ u⁴ du = u⁵/5 + C = tan⁵x / 5 + C

(e) ∫ x/(1 + x²) dx
Let u = 1 + x² ⇒ du = 2x dx ⇒ x dx = du/2
= ∫ (1/u)(du/2) = (1/2) ln|u| + C = (1/2) ln(1 + x²) + C

(f) ∫ cos²x sin x dx
Let u = cos x ⇒ du = –sin x dx ⇒ sin x dx = –du
= ∫ u² (–du) = –u³/3 + C = –cos³x / 3 + C

(g) ∫ –x² / √(1 + x³) dx
Let u = 1 + x³ ⇒ du = 3x² dx ⇒ x² dx = du/3
So integral = –∫ (1/√u)(du/3) = –(1/3) ∫ u⁻¹⁄² du = –(1/3)·2u¹⁄² + C = –(2/3)√(1 + x³) + C

(h) ∫ x√(1 + x²) dx
Let u = 1 + x² ⇒ du = 2x dx ⇒ x dx = du/2
= ∫ √u · (du/2) = (1/2) ∫ u¹⁄² du = (1/2)·(2/3)u³⁄² + C = (1/3)(1 + x²)^(3/2) + C

(i) ∫ e^√x / √x dx
Let u = √x = x¹⁄² ⇒ du = (1/(2√x)) dx ⇒ dx/√x = 2 du
Integral = ∫ eᵘ · 2 du = 2eᵘ + C = 2e^√x + C

(j) ∫ sec²(√x)/√x dx
Let u = √x ⇒ du = dx/(2√x) ⇒ dx/√x = 2 du
= ∫ sec²(u) · 2 du = 2 tan u + C = 2 tan(√x) + C

(k) ∫ 1/(x ln x) dx
Let u = ln x ⇒ du = dx/x
= ∫ (1/u) du = ln|u| + C = ln|ln x| + C

(l) ∫ (ln x)³ / x dx
Let u = ln x ⇒ du = dx/x
= ∫ u³ du = u⁴/4 + C = (ln x)⁴ / 4 + C

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3. Integration by parts

Recall: ∫ u dv = uv – ∫ v du

(a) ∫ x e²ˣ dx
Let u = x ⇒ du = dx
dv = e²ˣ dx ⇒ v = (1/2)e²ˣ
= x·(1/2)e²ˣ – ∫ (1/2)e²ˣ dx = (x/2)e²ˣ – (1/4)e²ˣ + C = e²ˣ (x/2 – 1/4) + C

(b) ∫ x cos x dx
u = x, du = dx; dv = cos x dx, v = sin x
= x sin x – ∫ sin x dx = x sin x + cos x + C

(c) ∫ xⁿ ln x dx
u = ln x, du = dx/x; dv = xⁿ dx, v = xⁿ⁺¹/(n+1) (n ≠ –1)
= ln x · xⁿ⁺¹/(n+1) – ∫ [xⁿ⁺¹/(n+1)]·(1/x) dx
= xⁿ⁺¹ ln x/(n+1) – (1/(n+1)) ∫ xⁿ dx
= xⁿ⁺¹ ln x/(n+1) – xⁿ⁺¹/[(n+1)²] + C
= xⁿ⁺¹/(n+1) [ln x – 1/(n+1)] + C

(d) ∫ eˣ cos x dx
Use integration by parts twice, or known formula:
Let I = ∫ eˣ cos x dx
First: u = cos x, dv = eˣ dx → du = –sin x dx, v = eˣ
I = eˣ cos x + ∫ eˣ sin x dx
Now J = ∫ eˣ sin x dx
u = sin x, dv = eˣ dx → du = cos x dx, v = eˣ
J = eˣ sin x – ∫ eˣ cos x dx = eˣ sin x – I
So I = eˣ cos x + eˣ sin x – I
⇒ 2I = eˣ (sin x + cos x)
⇒ I = (eˣ/2)(sin x + cos x) + C

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Final Answer:
1. (a) $ \frac{x^4}{4} + C $
(b) $ 6x^{1/3} + C $
(c) $ 2\ln|x| + C $
(d) $ \frac{1}{2}e^{2x} + C $
(e) $ \frac{5}{3}e^{3x/5} + C $
(f) $ -e^{-x} + C $
(g) $ \frac{1}{5}\sin(5x) + C $
(h) $ \frac{1}{2}\sin(2x) + C $
(i) $ 3\sin\left(\frac{x}{3}\right) + C $
(j) $ -\frac{1}{4}\cos(4x) + C $
(k) $ \frac{1}{2}\cos(2x) + C $
(l) $ \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C $
(m) $ \frac{1}{4}\arctan\left(\frac{x}{4}\right) + C $
(n) $ \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + C $
(o) $ \arcsin\left(\frac{x}{2}\right) + C $
(p) $ \arcsin\left(\frac{x}{3}\right) + C $
(q) $ \arcsin\left(\frac{x}{\sqrt{3}}\right) + C $

2. (a) $ -\frac{1}{3}\cos(x^3 + 7) + C $
(b) $ \frac{(1 + \sin x)^4}{4} + C $
(c) $ \frac{\sin^5 x}{5} + C $
(d) $ \frac{\tan^5 x}{5} + C $
(e) $ \frac{1}{2}\ln(1 + x^2) + C $
(f) $ -\frac{\cos^3 x}{3} + C $
(g) $ -\frac{2}{3}\sqrt{1 + x^3} + C $
(h) $ \frac{1}{3}(1 + x^2)^{3/2} + C $
(i) $ 2e^{\sqrt{x}} + C $
(j) $ 2\tan(\sqrt{x}) + C $
(k) $ \ln|\ln x| + C $
(l) $ \frac{(\ln x)^4}{4} + C $

3. (a) $ e^{2x}\left(\frac{x}{2} - \frac{1}{4}\right) + C $
(b) $ x\sin x + \cos x + C $
(c) $ \frac{x^{n+1}}{n+1}\left(\ln x - \frac{1}{n+1}\right) + C $
(d) $ \frac{e^x}{2}(\sin x + \cos x) + C $