Collection of 30 definite integrals from a) to z), showcasing a range of mathematical expressions involving trigonometric, logarithmic, exponential, and rational functions.

Let’s solve each integral one by one. I’ll go step by step for each, using simple methods like substitution or known formulas. We’ll check our work as we go.

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a) ∫₀¹ (x√(x³) – 1) dx
First, simplify: x√(x³) = x * x^(3/2) = x^(5/2)
So integral becomes: ∫₀¹ (x^(5/2) – 1) dx
Antiderivative: (2/7)x^(7/2) – x
Evaluate from 0 to 1: [(2/7)(1) – 1] – [0] = 2/7 – 1 = –5/7

✔ Check: Power rule applied correctly. Limits correct.

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b) ∫₀^(π/2) sin x cos x dx
Use identity: sin x cos x = (1/2) sin(2x)
Integral: (1/2) ∫₀^(π/2) sin(2x) dx
Antiderivative: (1/2)(–1/2 cos(2x)) = –1/4 cos(2x)
At π/2: –1/4 cos(π) = –1/4 (-1) = 1/4
At 0: –1/4 cos(0) = –1/4 (1) = –1/4
Result: 1/4 – (–1/4) = 1/2

✔ Alternative: Let u = sin x → du = cos x dx → ∫ u du from 0 to 1 = [u²/2]₀¹ = 1/2 — same answer.

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c) ∫₀^(e–1) ln(1+x) dx
Let u = 1 + x → du = dx; when x=0, u=1; x=e–1, u=e
Integral: ∫₁^e ln(u) du
Antiderivative of ln u is u ln u – u
At e: e ln e – e = e(1) – e = 0
At 1: 1 ln 1 – 1 = 0 – 1 = –1
Result: 0 – (–1) = 1

✔ Standard integral. Correct limits.

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d) ∫₀^(√(3π)/2) 4x cos(x²) dx
Let u = x² → du = 2x dx → 2 du = 4x dx
When x=0, u=0; x=√(3π)/2 → u = (3π)/4
Integral: ∫₀^(3π/4) 2 cos u du = 2 sin u |₀^(3π/4) = 2[sin(3π/4) – sin(0)] = 2[√2/2 – 0] = √2

✔ Substitution correct. sin(3π/4) = √2/2.

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e) ∫_(1/2)^1 x² ln x dx
Integration by parts: let u = ln x, dv = x² dx → du = 1/x dx, v = x³/3
Formula: uv – ∫v du = (x³/3)ln x – ∫ (x³/3)(1/x) dx = (x³/3)ln x – (1/3)∫x² dx
= (x³/3)ln x – (1/9)x³
Evaluate from 1/2 to 1:

At 1: (1/3)(0) – 1/9 = –1/9
At 1/2: ((1/8)/3) ln(1/2) – (1/9)(1/8) = (1/24)(–ln 2) – 1/72
So total: [–1/9] – [–(ln 2)/24 – 1/72] = –1/9 + (ln 2)/24 + 1/72
Convert to 72 denominator: –8/72 + 1/72 + (ln 2)/24 = –7/72 + (ln 2)/24
Write as: (ln 2)/24 – 7/72

✔ Double-check arithmetic: yes.

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f) ∫₁^e 2x ln x dx
Again, integration by parts: u = ln x, dv = 2x dx → du = 1/x dx, v = x²
uv – ∫v du = x² ln x – ∫x²*(1/x) dx = x² ln x – ∫x dx = x² ln x – x²/2
From 1 to e:

At e: e² * 1 – e²/2 = e²/2
At 1: 1*0 – 1/2 = –1/2
Result: e²/2 – (–1/2) = e²/2 + 1/2 = (e² + 1)/2

✔ Correct.

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g) ∫₀⁵ x/(x²+5) dx
Let u = x² + 5 → du = 2x dx → (1/2)du = x dx
Limits: x=0 → u=5; x=5 → u=30
Integral: (1/2) ∫₅³⁰ du/u = (1/2)[ln|u|]₅³⁰ = (1/2)(ln 30 – ln 5) = (1/2) ln(6)
Answer: (1/2) ln 6

✔ Log rules applied correctly.

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h) ∫₀¹ 1/(eˣ + eˣ) dx
Note: eˣ + e⁻ˣ = 2 cosh x, but maybe easier to multiply numerator and denominator by eˣ:
→ ∫₀¹ eˣ / (e²ˣ + 1) dx
Let u = eˣ → du = eˣ dx; x=0 → u=1; x=1 → u=e
Integral: ∫₁^e du/(u² + 1) = arctan u |₁^e = arctan(e) – arctan(1) = arctan(e) – π/4

✔ Good substitution.

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i) ∫₀¹ 6ˣ / 2ˣ dx = ∫₀¹ (6/2)ˣ dx = ₀¹ 3ˣ dx
Antiderivative of aˣ is aˣ / ln a → 3ˣ / ln 3
From 0 to 1: (3/ln 3) – (1/ln 3) = 2 / ln 3

✔ Simple exponent rule.

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j) ∫₀^(π/4) tan x / cos² x dx
Note: tan x / cos² x = sin x / cos³ x
Or better: write as tan x sec² x
Let u = tan x → du = sec² x dx
x=0 → u=0; x=π/4 → u=1
Integral: ∫₀¹ u du = [u²/2]₀¹ = 1/2

✔ Perfect substitution.

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k) ∫₁^e (ln³ x)/x dx
Let u = ln x → du = dx/x; x=1 → u=0; x=e → u=1
Integral: ∫₀¹ u³ du = [u⁴/4]₀¹ = 1/4

✔ Straightforward.

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l) ∫₀² x e^(1–x²) dx
Let u = 1 – x² → du = –2x dx → x dx = –du/2
x=0 → u=1; x=2 → u=1–4=–3
Integral: ∫₁^(-3) eᵘ (–du/2) = (1/2) ∫_{-3}^1 eᵘ du = (1/2)(e¹ – e⁻³) = (e – e⁻³)/2

✔ Sign handled correctly.

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m) ∫₀^π x cos(2x) dx
Integration by parts: u = x, dv = cos(2x) dx → du = dx, v = (1/2) sin(2x)
uv – ∫v du = (x/2) sin(2x) – ∫ (1/2) sin(2x) dx = (x/2) sin(2x) + (1/4) cos(2x)
Evaluate 0 to π:

At π: (π/2) sin(2π) + (1/4) cos(2π) = 0 + 1/4
At 0: 0 + (1/4) cos(0) = 1/4
Result: 1/4 – 1/4 = 0

✔ Symmetry? Yes, function is odd around π/2? Anyway, calculation checks out.

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n) ∫₋₄^(-3) 1/(x² – 4) dx
Factor: x² – 4 = (x–2)(x+2) → partial fractions:
1/[(x–2)(x+2)] = A/(x–2) + B/(x+2)
Solve: 1 = A(x+2) + B(x–2)
Set x=2: 1 = A(4) → A=1/4
Set x=-2: 1 = B(-4) → B=–1/4
So integral: (1/4)∫ [1/(x–2) – 1/(x+2)] dx from –4 to –3
= (1/4)[ln|x–2| – ln|x+2|] from –4 to –3
= (1/4)[ln|(x–2)/(x+2)|] from –4 to –3

At –3: ln|(-5)/(-1)| = ln 5
At –4: ln|(-6)/(-2)| = ln 3
Result: (1/4)(ln 5 – ln 3) = (1/4) ln(5/3)

✔ Partial fractions done right.

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o) ∫₀¹ 2√x / (1+x) dx
Let u = √x → x = u², dx = 2u du
x=0 → u=0; x=1 → u=1
Integral: ∫₀¹ 2u / (1+u²) * 2u du = ∫₀¹ 4u² / (1+u²) du
Now: 4u²/(1+u²) = 4[1 – 1/(1+u²)]
So integral: 4 ∫₀¹ [1 – 1/(1+u²)] du = 4[u – arctan u]₀¹ = 4[(1 – π/4) – 0] = 4 – π

✔ Algebraic manipulation correct.

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p) ∫₁² 1/[x √(1 – ln² x)] dx
Let u = ln x → du = dx/x; x=1 → u=0; x=2 → u=ln 2
Integral: ∫₀^(ln 2) 1/√(1 – u²) du = arcsin u |₀^(ln 2) = arcsin(ln 2)

✔ Standard form. Note: ln 2 ≈ 0.693 < 1, so valid.

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q) ∫_(π/4)^(π/3) 1/cos² x dx = ∫ sec² x dx = tan x |_(π/4)^(π/3) = tan(π/3) – tan(π/4) = √3 – 1
Answer: √3 – 1

✔ Basic derivative.

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r) ∫₋₂² 6/(8 + 3x²) dx
Factor out constants: 6 ∫₋₂² dx/(8 + 3x²) = 6 ∫₂² dx/[3(x² + 8/3)] = 2 ∫₋₂² dx/(x² + (√(8/3))²)
Standard form: ∫ dx/(x² + a²) = (1/a) arctan(x/a)
Here a = √(8/3) = 2√(2/3)
So integral: 2 * [1/a arctan(x/a)] from –2 to 2
Since even function: 2 * 2 * [1/a arctan(2/a)] = 4/a arctan(2/a)

Compute a = √(8/3) = 2√2 / √3
So 4 / (2√2 / √3) = 4 * √3 / (2√2) = 2 √3 / 2 = √6 * √2 * √3 / √2? Wait, better:

2 √3 / 2 = 2 (3/2) = 2 √6 / 2 = √6? No:

Actually: 2 √3 / 2 = 2 (3/2) = 2 √6 / 2? Multiply num and den by √2:
2 √3 √2 / 2 = √6? Wait no:

2 √3 / 2 = (2/√2) √3 = 2 * √3 = √6? Yes! Because 2/√2 = √2.

So coefficient is √6
Then arctan(2 / a) = arctan(2 / (√(8/3))) = arctan(2 * √(3/8)) = arctan(√(4 * 3 / 8)) = arctan(√(12/8)) = arctan(√(3/2))

Wait, let me recast:

Original: 2 ∫₋₂² dx/(x² + 8/3) = 2 * [1/√(8/3) arctan(x / √(8/3))] from –2 to 2
= 2 * [1/√(8/3)] * 2 arctan(2 / √(8/3)) [since even]

√(8/3) = √8 / √3 = 2√2 / √3
So 1/√(8/3) = √3 / (2√2)
Thus: 2 * [√3 / (2√2)] * 2 arctan(2 * √3 / (2√2)) = [2 √3 / 2] arctan(√3 / √2)
Simplify: 2 √(3/2) arctan(√(3/2)) = √6 * 2 / √2? Messy.

Better numerical approach? Or leave symbolic.

Actually, standard way:

∫ dx/(a² + x²) = (1/a) arctan(x/a)

We have ∫ dx/(8/3 + x²) = ∫ dx/( (√(8/3))² + x² ) = (1/√(8/3)) arctan(x / √(8/3))

So overall: 2 * [that] from –2 to 2 = 2 * 2 * (1/√(8/3)) arctan(2 / √(8/3)) = 4 / √(8/3) * arctan(2 √(3/8))

Simplify 4 / √(8/3) = 4 √(3/8) = 4 √3 / 8 = 4 √3 / (2√2) = 2 √3 / √2 = √6 * √2 * √3 / √2? Again:

2 √3 / 2 = √4 * √3 / √2 = √(12/2) = √6? No: (2 √3)/√2 = 2 √(3/2) = √4 * √(3/2) = √6

Yes! So coefficient is √6

And argument: 2 / √(8/3) = 2 √(3/8) = √(4 * 3 / 8) = √(12/8) = √(3/2)

So answer: √6 arctan(√(3/2))

But perhaps simplify differently.

Note: 8 + 3x² = 3(x² + 8/3), so integral is 6 ∫ dx/(3(x² + 8/3)) = 2 ∫ dx/(x² + 8/3)

As above.

I think it's fine to leave as: 2 * [arctan(x / √(8/3)) / √(8/3)] from -2 to 2, but since symmetric, 4 / √(8/3) * arctan(2 / √(8/3))

Numerically: √(8/3) ≈ √2.666 ≈ 1.633, 2/1.633≈1.224, arctan(1.224)≈0.888 rad, 4/1.633≈2.45, times 0.888≈2.17 — but we want exact.

Alternatively, factor differently:

6 / (8 + 3x²) = 6 / [3(x² + 8/3)] = 2 / (x² + 8/3)

Same as before.

I recall that ∫ dx/(a² + b²x²) = (1/(ab)) arctan(bx/a)

Here, 8 + 3x² = 3x² + 8, so a²=8, b²=3? Not quite.

Write as ∫ dx / ( (√8)^2 + (√3 x)^2 ) = (1/(√8 √3)) arctan( (√3 x)/√8 ) ? Let's see:

General: ∫ dx/(A + B x²) = (1/√(AB)) arctan( x √(B/A) ) if A,B>0

Here A=8, B=3, so ∫ dx/(8+3x²) = (1/√(24)) arctan( x √(3/8) )

Then multiplied by 6: 6 * (1/√24) arctan( x √(3/8) ) from –2 to 2

√24 = 2√6, so 6/(2√6) = 3/√6 = √6 / 2

Then times 2 (because from –2 to 2, and odd function inside? No, arctan is odd, but evaluated at ±2, so difference is 2 arctan(...)

So: [√6 / 2] * 2 arctan(2 √(3/8)) = √6 arctan(2 √(3/8))

Simplify 2 √(3/8) = 2 √3 / √8 = 2 √3 / (2√2) = √3 / √2 = √(3/2)

So finally: √6 arctan(√(3/2))

✔ Acceptable exact form.

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s) ∫₀¹ x / √(4 – x²) dx
Let u = 4 – x² → du = –2x dx → x dx = –du/2
x=0 → u=4; x=1 → u=3
Integral: ∫³ (–du/2) / √u = (1/2) ∫₃⁴ u^{-1/2} du = (1/2) [2 u^{1/2}]₃⁴ = [√u]₃⁴ = √4 – 3 = 2 – √3

✔ Simple substitution.

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t) ∫₀^(1/2) 2(1+x²)/(1–x²) dx
Note: (1+x²)/(1–x²) = [ (1–x²) + 2x² ] / (1–x²) = 1 + 2x²/(1–x²)
But better: divide directly or use polynomial division.

(1+x²)/(1–x²) = – (x²+1)/(x²–1) = – [ (x²–1) + 2 ] / (x²–1) = – [1 + 2/(x²–1)] = –1 – 2/(x²–1)

But x²–1 = –(1–x²), so –2/(x²–1) = 2/(1–x²)

So (1+x²)/(1–x²) = –1 + 2/(1–x²)? Let me compute:

Set f(x) = (1+x²)/(1–x²)

Write as: [ -(1–x²) + 2 ] / (1–x²) = -1 + 2/(1–x²)

Yes! Because –(1–x²) = –1 + x², then +2 gives 1 + x². Perfect.

So integral: 2 ∫₀^(1/2) [ –1 + 2/(1–x²) ] dx = 2 [ –x + 2 ∫ dx/(1–x²) ]

Now ∫ dx/(1–x²) = (1/2) ln|(1+x)/(1–x)| by partial fractions.

So overall: 2 [ –x + 2 * (1/2) ln|(1+x)/(1–x)| ] = 2 [ –x + ln|(1+x)/(1–x)| ]

From 0 to 1/2:

At 1/2: –1/2 + ln|(1.5)/(0.5)| = –0.5 + ln 3
At 0: 0 + ln(1) = 0
So inside: (–0.5 + ln 3) – 0 = ln 3 – 0.5
Multiply by 2: 2 ln 3 – 1

✔ Correct.

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u) ∫₀^π sin x cos² x dx
Let u = cos x → du = –sin x dx
x=0 → u=1; x=π → u=–1
Integral: ∫₁^(-1) u² (–du) = ∫_{-1}^1 u² du = 2 ∫₀¹ u² du = 2 [u³/3]₀¹ = 2/3
Answer: 2/3

✔ Even function, symmetric.

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v) ∫₀¹ x e^{-x} dx
Integration by parts: u = x, dv = e^{-x} dx → du = dx, v = –e^{-x}
uv – ∫v du = –x e^{-x} – ∫ –e^{-x} dx = –x e^{-x} – e^{-x}
From 0 to 1:

At 1: –1*e^{-1} – e^{-1} = –2/e
At 0: –0 – 1 = –1
Result: (–2/e) – (–1) = 1 – 2/e

✔ Standard.

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w) ∫₀¹ 2 √(2x + x²) dx
Complete square: 2x + x² = x² + 2x = (x+1)^2 – 1
So √((x+1)^2 – 1)
Let u = x+1 → du = dx; x=0 → u=1; x=1 → u=2
Integral: 2 ∫₁² √(u² – 1) du

This is standard: ∫ √(u² – a²) du = (u/2) √(u² – a²) – (a²/2) ln|u + √(u² – a²)|
Here a=1, so:

= 2 [ (u/2) √(u² – 1) – (1/2) ln|u + √(u² – 1)| ] from 1 to 2

At u=2: (2/2)√(4–1) – (1/2) ln(2 + √3) = √3 – (1/2) ln(2+√3)
At u=1: (1/2)√0 – (1/2) ln(1+0) = 0 – 0 = 0
So inside: √3 – (1/2) ln(2+√3)
Multiply by 2: 2√3 – ln(2+√3)

✔ Correct formula application.

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x) ∫_(π/4)^(π/2) x / sin² x dx = ∫ x csc² x dx
Integration by parts: u = x, dv = csc² x dx → du = dx, v = –cot x
uv – ∫v du = –x cot x – ∫ –cot x dx = –x cot x + ∫ cot x dx = –x cot x + ln|sin x|
From π/4 to π/2:

At π/2: –(π/2)*0 + ln(sin π/2) = 0 + ln 1 = 0
At π/4: –(π/4)*1 + ln(sin π/4) = –π/4 + ln(√2/2) = –π/4 + ln(2^{-1/2}) = –π/4 – (1/2) ln 2
Result: 0 – [–π/4 – (1/2) ln 2] = π/4 + (1/2) ln 2

✔ Careful with signs.

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y) ∫₀¹ arctan x dx
Integration by parts: u = arctan x, dv = dx → du = 1/(1+x²) dx, v = x
uv – ∫v du = x arctan x – ∫ x/(1+x²) dx
∫ x/(1+x²) dx = (1/2) ln(1+x²)
So: x arctan x – (1/2) ln(1+x²) from 0 to 1

At 1: 1*(π/4) – (1/2) ln 2 = π/4 – (1/2) ln 2
At 0: 0 – 0 = 0
Result: π/4 – (1/2) ln 2

✔ Standard.

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z) ∫₀^(π/2) e^{2x} cos x dx
This requires integration by parts twice or use formula.

Recall: ∫ e^{ax} cos bx dx = e^{ax}/(a²+b²) (a cos bx + b sin bx) + C

Here a=2, b=1
So antiderivative: e^{2x}/(4+1) (2 cos x + sin x) = (e^{2x}/5)(2 cos x + sin x)

From 0 to π/2:

At π/2: (e^π / 5)(2*0 + 1) = e^π / 5
At 0: (1/5)(2*1 + 0) = 2/5
Result: e^π / 5 – 2/5 = (e^π – 2)/5

✔ Formula applied correctly.

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Z) ∫₀^(π/2) 4 sin x cos³ x dx
Let u = cos x → du = –sin x dx
x=0 → u=1; x=π/2 → u=0
Integral: 4 ∫₁⁰ u³ (–du) = 4 ∫₀¹ u³ du = 4 [u⁴/4]₀¹ = 4*(1/4) = 1

✔ Very clean.

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Now, compiling all answers:

a) –5/7
b) 1/2
c) 1
d) √2
e) (ln 2)/24 – 7/72
f) (e² + 1)/2
g) (1/2) ln 6
h) arctan(e) – π/4
i) 2 / ln 3
j) 1/2
k) 1/4
l) (e – e⁻³)/2
m) 0
n) (1/4) ln(5/3)
o) 4 – π
p) arcsin(ln 2)
q) √3 – 1
r) √6 arctan(√(3/2))
s) 2 – √3
t) 2 ln 3 – 1
u) 2/3
v) 1 – 2/e
w) 2√3 – ln(2+√3)
x) π/4 + (1/2) ln 2
y) π/4 – (1/2) ln 2
z) (e^π – 2)/5
Z) 1

Final Answer:
a) -\frac{5}{7}
b) \frac{1}{2}
c) 1
d) \sqrt{2}
e) \frac{\ln 2}{24} - \frac{7}{72}
f) \frac{e^2 + 1}{2}
g) \frac{1}{2} \ln 6
h) \arctan(e) - \frac{\pi}{4}
i) \frac{2}{\ln 3}
j) \frac{1}{2}
k) \frac{1}{4}
l) \frac{e - e^{-3}}{2}
m) 0
n) \frac{1}{4} \ln \frac{5}{3}
o) 4 - \pi
p) \arcsin(\ln 2)
q) \sqrt{3} - 1
r) \sqrt{6} \arctan \sqrt{\frac{3}{2}}
s) 2 - \sqrt{3}
t) 2 \ln 3 - 1
u) \frac{2}{3}
v) 1 - \frac{2}{e}
w) 2\sqrt{3} - \ln(2 + \sqrt{3})
x) \frac{\pi}{4} + \frac{1}{2} \ln 2
y) \frac{\pi}{4} - \frac{1}{2} \ln 2
z) \frac{e^\pi - 2}{5}
Z) 1