Topic 01
Integration by Parts
⚡ Instant Memory — IBP
LIATE
Choose u in this order: Logarithm → Inverse trig → Algebraic → Trig → Exponential. The other factor becomes dv.
FORMULA
∫u dv = uv − ∫v du | Always differentiate the LIATE choice.
TRAP
For ∫arcsin(x) dx: let u = arcsin(x), dv = dx. Then √(1−x²) appears — don't forget to simplify!
Q01
EASY
Find \(\displaystyle\int x e^x\, dx\)
hint · LIATE — which is u?
Step-by-Step Solution
①LIATE → u = x (Algebraic), dv = eˣ dx
②du = dx, v = eˣ
③∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C
$$\int x e^x\,dx = xe^x - e^x + C$$
Q02
EASY
Find \(\displaystyle\int x\ln x\, dx\)
hint · ln x is LIATE-first!
Step-by-Step Solution
①u = ln x → du = 1/x dx; dv = x dx → v = x²/2
②= (x²/2)ln x − ∫(x²/2)(1/x) dx = (x²/2)ln x − ∫(x/2) dx
③= (x²/2)ln x − x²/4 + C
$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$
Q03
MEDIUM
Evaluate \(\displaystyle\int_0^{4/5} \arcsin(x)\, dx\)
hint · use sin(arcsin 4/5)=4/5, cos=3/5
Step-by-Step Solution
①u = arcsin x, dv = dx → du = dx/√(1−x²), v = x
②= [x arcsin x]₀^{4/5} − ∫₀^{4/5} x/√(1−x²) dx
③∫x/√(1−x²) dx = −√(1−x²) + C (sub w=1−x²)
④= (4/5)arcsin(4/5) − [−√(1−x²)]₀^{4/5} = (4/5)arcsin(4/5) − (−3/5+1) = (4/5)arcsin(4/5) − 2/5
$$\int_0^{4/5}\!\arcsin x\,dx = \frac{4}{5}\arcsin\!\frac{4}{5} - \frac{2}{5}$$
Q04
MEDIUM
Find \(\displaystyle\int x^2 \cos x\, dx\)
hint · apply IBP twice!
Step-by-Step Solution (IBP twice)
①Round 1: u=x², dv=cos x dx → v=sin x. Get x²sin x − 2∫x sin x dx
②Round 2: ∫x sin x dx. u=x, dv=sin x dx → = −x cos x + ∫cos x dx = −x cos x + sin x
③Combine: x²sin x − 2(−x cos x + sin x) + C = x²sin x + 2x cos x − 2sin x + C
$$x^2\sin x + 2x\cos x - 2\sin x + C$$
Topic 02
U-Substitution
⚡ Instant Memory — Substitution
SPOT IT
Look for f(g(x))·g'(x). The "inside function" is your u.
BOUNDS
For definite integrals: always change the bounds using the u-formula. Never substitute back after integrating!
TRICK
If you see sinˢ·cosᵗ with odd power: save one factor, convert rest via sin²+cos²=1, then sub u=cos or u=sin.
Q05
EASY
Find \(\displaystyle\int \frac{2x}{x^2+1}\, dx\)
hint · what's the derivative of x²+1?
Step-by-Step Solution
①Let u = x²+1 → du = 2x dx ✓ (numerator is exactly du!)
②∫(2x dx)/(x²+1) = ∫du/u = ln|u| + C
③Back-sub: ln(x²+1) + C (x²+1 > 0 always)
Q06
EASY
Evaluate \(\displaystyle\int_0^{\arctan(1/2)}\!(\sin x)(\cos x)\, dx\)
hint · let u = sin x, OR note sin x cos x = ½sin(2x)
Step-by-Step Solution
①u = sin x → du = cos x dx. Bounds: x=0 → u=0; x=arctan(1/2) → sin(arctan 1/2)=1/√5
②= ∫₀^{1/√5} u du = [u²/2]₀^{1/√5} = 1/(2·5) = 1/10
Answer: \(\dfrac{1}{10}\)
Q07
EASY
Find \(\displaystyle\int e^{3x+1}\, dx\)
hint · u = 3x+1
Step-by-Step Solution
①u = 3x+1 → du = 3dx → dx = du/3
②∫eᵘ (du/3) = (1/3)eᵘ + C = (1/3)e^{3x+1} + C
Topic 03
Trig Integrals
⚡ Instant Memory — Trig Integrals
ODD POWER
sinⁿx with n odd → save one sinx, convert rest: sin²x = 1−cos²x, then u=cosx.
EVEN POWER
Use half-angle: sin²x = (1−cos2x)/2, cos²x = (1+cos2x)/2.
∫tanx
= ln|sec x| + C | ∫sec x = ln|sec x + tan x| + C — these must be memorized!
Q08
MEDIUM
Evaluate \(\displaystyle\int_0^{\arcsin(3/5)}\! 3\sin^3\!x\, dx\)
hint · sin³x = sinx(1−cos²x); at arcsin(3/5), cosx=4/5
Step-by-Step Solution
①3∫sin³x dx = 3∫sinx(1−cos²x)dx. Let u=cosx, du=−sinx dx
②= 3∫(1−u²)(−du) = 3[−u + u³/3] + C = 3[−cosx + cos³x/3]
③At x=arcsin(3/5): cosx=4/5. At x=0: cosx=1.
④= 3[(−4/5 + 64/375) − (−1+1/3)] = 3[(−300+64)/375 + 2/3] = 3[−236/375 + 250/375] = 3(14/375) = 42/375 = 14/125
$$3\int_0^{\arcsin(3/5)}\!\sin^3 x\,dx = \frac{14}{125}$$
Q09
EASY
Find \(\displaystyle\int \cos^2\!x\, dx\)
hint · half-angle identity!
Step-by-Step Solution
①cos²x = (1 + cos 2x)/2
②∫cos²x dx = ∫(1+cos2x)/2 dx = x/2 + (sin2x)/4 + C
Q10
EASY
Find \(\displaystyle\int \tan x\, dx\)
hint · write tan = sin/cos, then sub u = cos x
Step-by-Step Solution
①∫tan x dx = ∫(sin x / cos x) dx. Let u = cos x → du = −sin x dx
②= ∫(−du)/u = −ln|u| + C = −ln|cos x| + C (= ln|sec x| + C)
Topic 04
Trigonometric Substitution
⚡ Instant Memory — Trig Sub
√(a²−x²)
x = a sinθ → √(a²−x²) = a cosθ. Draw a right triangle!
√(a²+x²)
x = a tanθ → √(a²+x²) = a secθ.
√(x²−a²)
x = a secθ → √(x²−a²) = a tanθ.
Q11
EASY
Find \(\displaystyle\int \frac{1}{\sqrt{1-x^2}}\, dx\)
hint · this is a standard formula — do you know it?
Step-by-Step Solution
①Let x = sinθ → dx = cosθ dθ, √(1−x²) = cosθ
②∫cosθ dθ / cosθ = ∫dθ = θ + C = arcsin x + C
Standard formula: \(\displaystyle\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C\)
Q12
HARD
Evaluate \(\displaystyle\int_1^{5/3}\!\sqrt{x^2-1}\, dx\)
hint · formula ∫√(x²−1)dx = (x√(x²−1))/2 − (1/2)ln|x+√(x²−1)|
Step-by-Step Solution
①Use: ∫√(x²−1) dx = (x/2)√(x²−1) − (1/2)ln|x+√(x²−1)| + C
②At x=5/3: √(25/9−1) = √(16/9) = 4/3. Value = (5/3)(4/3)/2 − (1/2)ln(5/3+4/3) = 20/18 − (1/2)ln3 = 10/9 − (1/2)ln3
③At x=1: √0=0, 1+0=1, ln1=0. Value = 0.
④Result = 10/9 − (1/2)ln3
$$\int_1^{5/3}\!\sqrt{x^2-1}\,dx = \frac{10}{9} - \frac{1}{2}\ln 3$$
Q13
EASY
Find \(\displaystyle\int \frac{1}{x^2+4}\, dx\)
hint · standard form: ∫dx/(x²+a²) = (1/a)arctan(x/a)
Step-by-Step Solution
①Write x²+4 = 4(x²/4+1) = 4((x/2)²+1)
②∫dx/(x²+4) = (1/4)∫dx/((x/2)²+1). Let u=x/2 → du=dx/2
③= (1/4)·2∫du/(u²+1) = (1/2)arctan(x/2) + C
Topic 05
Improper Integrals
⚡ Instant Memory — Improper Integrals
LIMIT
∫ₐ^∞ f(x)dx = lim_{t→∞} ∫ₐᵗ f(x)dx. Always replace ∞ with a limit — never plug in directly.
P-TEST
∫₁^∞ 1/xᵖ dx converges iff p > 1. ∫₀¹ 1/xᵖ dx converges iff p < 1.
∞ TRAP
e^{−x} → 0 as x→∞. ln x → ∞ but slow. Watch which beats which!
Q14
EASY
Evaluate \(\displaystyle\int_1^{\infty} \frac{1}{x^2}\, dx\)
hint · p-test: p=2 > 1 → converges
Step-by-Step Solution
①= lim_{t→∞} [−1/x]₁ᵗ = lim_{t→∞} (−1/t + 1) = 0 + 1 = 1
Q15
HARD
Evaluate \(\displaystyle\int_{\ln 4}^{\infty}\!\frac{1}{\sqrt{9+e^{2x}}}\, dx\)
hint · sub u=eˣ, then ∫du/(u√(u²+9)) = −(1/3)ln|(3+√(u²+9))/u|
Step-by-Step Solution
①Let u = eˣ → du = eˣ dx → dx = du/u. Bounds: x=ln4 → u=4; x→∞ → u→∞
②= ∫₄^∞ du/(u√(u²+9)). Use formula: ∫du/(u√(u²+a²)) = −(1/a)ln|(a+√(u²+a²))/u| + C, a=3
③= [−(1/3)ln|(3+√(u²+9))/u|]₄^∞. As u→∞: (3+u)/u→1 → ln→0. At u=4: √25=5, (3+5)/4=2.
④= −(1/3)[0 − ln 2] = (ln2)/3
$$\int_{\ln 4}^{\infty}\!\frac{dx}{\sqrt{9+e^{2x}}} = \frac{\ln 2}{3}$$
Q16
EASY
Does \(\displaystyle\int_1^{\infty}\frac{1}{x}\, dx\) converge or diverge?
hint · classic harmonic series test!
Step-by-Step Solution
①= lim_{t→∞} [ln x]₁ᵗ = lim_{t→∞} (ln t − 0) = ∞. Diverges.
②P-test: p=1 → NOT p>1 → diverges. This is the borderline case!
Topic 06
Advanced Combinations
⚡ Instant Memory — Advanced
KING RULE
∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b−x)dx. Powerful when f(x)+f(a+b−x) simplifies to a constant!
PARTIAL FRAC
For rational P(x)/Q(x) with deg P < deg Q: factor denominator, decompose into A/(x−r) + B/(x−s) …
CYCLIC IBP
∫eˣsin x dx: apply IBP twice → original integral reappears. Solve algebraically!
Q17
MEDIUM
Find \(\displaystyle\int e^x \sin x\, dx\)
hint · IBP twice, then collect the integral from both sides
Step-by-Step Solution (Cyclic IBP)
①I = ∫eˣ sin x dx. IBP: u=sin x, dv=eˣ dx → I = eˣ sin x − ∫eˣ cos x dx
②IBP again: ∫eˣ cos x dx. u=cos x → = eˣ cos x + ∫eˣ sin x dx = eˣ cos x + I
③So: I = eˣ sin x − (eˣ cos x + I) → 2I = eˣ(sin x − cos x) → I = ½eˣ(sin x − cos x) + C
Q18
MEDIUM
Find \(\displaystyle\int \frac{1}{x^2-1}\, dx\)
hint · partial fractions: 1/(x²−1) = A/(x−1) + B/(x+1)
Step-by-Step Solution
①1/(x²−1) = 1/((x−1)(x+1)) = A/(x−1) + B/(x+1)
②Solve: A=1/2, B=−1/2
③∫ = (1/2)ln|x−1| − (1/2)ln|x+1| + C = (1/2)ln|(x−1)/(x+1)| + C
Q19
HARD
Using the King Rule, evaluate \(\displaystyle\int_0^{\pi}\!\frac{x\sin x}{1+\cos^2\!x}\, dx\)
hint · let I = ∫, then I = ∫(π−x)sin x/(1+cos²x)dx. Add both!
Step-by-Step Solution (King Rule)
①Let I = ∫₀^π x sinx/(1+cos²x) dx. King rule (x→π−x): sin(π−x)=sinx, cos(π−x)=−cosx, cos²(π−x)=cos²x
②So I = ∫₀^π (π−x)sinx/(1+cos²x) dx
③Add: 2I = π∫₀^π sinx/(1+cos²x) dx. Let u=cosx → = π∫₁^{-1} (−du)/(1+u²) = π[arctan u]_{-1}^{1} = π·(π/4−(−π/4)) = π²/2
④Therefore I = π²/4
$$\int_0^{\pi}\!\frac{x\sin x}{1+\cos^2 x}\,dx = \frac{\pi^2}{4}$$
Q20
EASY
Find \(\displaystyle\int \frac{\ln x}{x}\, dx\)
hint · super simple sub — what's d(ln x)?
Step-by-Step Solution
①Let u = ln x → du = dx/x. Notice (ln x)(dx/x) = u du
②∫u du = u²/2 + C = (ln x)²/2 + C