NDA Definite Integrals previous year questions with solutions

Answer: (d) 1c(a + b + c)

Method. ∫ dx(px + c)² = -1p(px + c) with p = a + b

ax + bx + c = (a + b)x + c; write p = a + b.
∫₀¹ dx(px + c)² = [-1p(px + c)]₀¹ = 1p(1c − 1p + c) = 1p · pc(p + c).
= 1c(a + b + c).

Answer: (b) 3

Method. the integrand is the derivative of x^x

y = x^x ⇒ ln y = x ln x ⇒ y'y = ln x + 1, so ddxx^x = x^x(1 + ln x).
∫₁² x^x(1 + ln x) dx = [x^x]₁² = 2² - 1¹ = 3.

Answer: (a) I₁ - I₂ = 0

Method. substitution x = e^t in I₁

In I₁ put x = e^t: dx = e^t dt, ln x = t; x: e → e² gives t: 1 → 2.
I₁ = ∫₁² e^tt dt, which is I₂ (the variable name does not matter).
So I₁ - I₂ = 0.

Answer: (c) ln2

Method. half-angle substitution t = tanx2

Put t = tanx2: sin x = 2t1 + t², cos x = 1 − t²1 + t², dx = 2dt1 + t²; x: 0 → π2 gives t: 0 → 1.
g(x) = 1 + sin x + cos x = (1 + t²) + 2t + (1 − t²)1 + t² = 2(1 + t)1 + t².
∫₀^π/⁠2 dxg(x) = ∫₀¹ 1 + t²2(1 + t) · 2dt1 + t² = ∫₀¹ dt1 + t = [ln(1 + t)]₀¹ = ln 2.

Answer: (c) π4 − ln22

Method. property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx, then the previous item

Put x → π2 − x: g is unchanged, so I = ∫₀^π/⁠2 cos xg(x) dx as well.
Adding: 2I = ∫₀^π/⁠2 sin x + cos x1 + sin x + cos x dx = ∫₀^π/⁠2 (1 − 1g(x)) dx = π2 − ln 2 (previous item).
I = π4 − ln 22.

Answer: (b) 2

Method. form three linear equations in P, Q, R from the three conditions and solve

f(0) = P + Q + R = 6.
f'(x) = Pe^x + 2Qe^2x + 3Re^3x; with e^{ln 3} = 3: f'(ln 3) = 3P + 18Q + 81R = 282, i.e. P + 6Q + 27R = 94.
∫₀^ln2 f(x)dx = P(2 −⁠ 1) + Q2(4 − 1) + R3(8 − 1) = P + 3Q2 + 7R3 = 11, i.e. 6P + 9Q + 14R = 66.
Eliminate P = 6 − Q − R: 5Q + 26R = 88 and 3Q + 8R = 30.
From these, 3(5Q + 26R) − 5(3Q + 8R) = 264 − 150, so 38R = 114, R = 3; then Q = 30 − 243 = 2 and P = 6 − 2 − 3 = 1.
Check: 1 + 2 + 3 = 6; 3 + 36 + 243 = 282; 1 + 3 + 7 = 11.
Q = 2.
Answer: option (b).

Answer: (c) 3

Method. same linear system in P, Q, R as the previous item

f(0) = P + Q + R = 6.
f'(x) = Pe^x + 2Qe^2x + 3Re^3x; with e^{ln 3} = 3: f'(ln 3) = 3P + 18Q + 81R = 282, i.e. P + 6Q + 27R = 94.
∫₀^ln2 f(x)dx = P(2 −⁠ 1) + Q2(4 − 1) + R3(8 − 1) = P + 3Q2 + 7R3 = 11, i.e. 6P + 9Q + 14R = 66.
Eliminate P = 6 − Q − R: 5Q + 26R = 88 and 3Q + 8R = 30.
From these, 3(5Q + 26R) − 5(3Q + 8R) = 264 − 150, so 38R = 114, R = 3; then Q = 30 − 243 = 2 and P = 6 − 2 − 3 = 1.
R = 3.
Answer: option (c).

Answer: (d) 14

Method. solve for P, Q, R, then f'(0) = P + 2Q + 3R

f(0) = P + Q + R = 6.
f'(x) = Pe^x + 2Qe^2x + 3Re^3x; with e^{ln 3} = 3: f'(ln 3) = 3P + 18Q + 81R = 282, i.e. P + 6Q + 27R = 94.
∫₀^ln2 f(x)dx = P(2 −⁠ 1) + Q2(4 − 1) + R3(8 − 1) = P + 3Q2 + 7R3 = 11, i.e. 6P + 9Q + 14R = 66.
Eliminate P = 6 − Q − R: 5Q + 26R = 88 and 3Q + 8R = 30.
From these, 3(5Q + 26R) − 5(3Q + 8R) = 264 − 150, so 38R = 114, R = 3; then Q = 30 − 243 = 2 and P = 6 − 2 − 3 = 1.
f'(0) = P + 2Q + 3R = 1 + 4 + 9 = 14.
Answer: option (d).

Answer: (b) 12

Method. divide by cos² x and substitute t = 1 + tan x

(sin x + cos x)² = cos² x (tan x + 1)², so the integrand is sec² x(1 + tan x)².
Put t = 1 + tan x, dt = sec² x dx; x = 0 gives t = 1, x = π4 gives t = 2.
∫₁² dtt² = [-1t]₁² = -12 + 1 = 12.
Answer: option (b).

Answer: (b) 0

Method. fundamental theorem of calculus: ∫_a^b f'(x)dx = f(b) − f(a)

∫₁⁴ f'(x) dx = f(4) − f(1).
Given f(1) = f(4), this is 0.
Answer: option (b).

Answer: (c) 1

Method. e^{ln t} = t for t > 0

For 0 ≤ x < π2, cos x > 0, so e^{ln(cos x)} = cos x (the single point x = π2 does not affect the integral).
∫₀^π/⁠2 cos x dx = [sin x]₀^π/⁠2 = 1 − 0 = 1.
Answer: option (c).

Answer: (d) 8ln2

Method. f'(x)f(x) = ddx ln f(x); here ln 2^x = x ln 2

f(x) = 2^x, f'(x) = 2^x ln 2, so f'(x)f(x) = ln 2.
∫₂¹⁰ ln 2 dx = (10 − 2) ln 2 = 8 ln 2.
Answer: option (d).

Answer: (b) 12

Method. factor tan x(tan² x + 1) = tan x sec² x, then t = tan x

tan³ x + tan x = tan x(1 + tan² x) = tan x sec² x.
Put t = tan x, dt = sec² x dx; x = 0 gives t = 0, x = π4 gives t = 1.
∫₀¹ t dt = 12.
Answer: option (b).

Answer: (b) e − 1

Method. substitution t = sin x

Put t = sin x, dt = cos x dx; x = 0 gives t = 0, x = π2 gives t = 1.
∫₀^π/⁠2 e^{sin x} cos x dx = ∫₀¹ e^t dt = e − 1.
Answer: option (b).

Answer: (c) -π2

Method. differentiate first: ddxtan^-11x = -11 + x² for x ≠ 0

For x ≠ 0: ddxtan^-11x = 11 + 1x²·(-1x²) = -11 + x².
The integrand is -11 + x² (except at the single point x = 0), so
∫_-1¹ (-11 + x²) dx = -[tan^-1 x]_-1¹ = -(π4 + π4) = -π2.
Trap: writing [tan^-11x]_-1¹ = π4 − (-π4) = π2 is wrong, because tan^-11x jumps from -π2 to π2 at x = 0, so it is not an antiderivative on the whole of [-1, 1].
Answer: -π2, option (c).

Answer: (a) e^π + 12

Method. standard result ∫ e^x sin x dx = e^x2(sin x − cos x)

Integrating by parts twice, ∫ e^x sin x dx = e^x2(sin x − cos x) + C.
At x = π: e^π2(0 − (-1)) = e^π2.
At x = 0: 12(0 − 1) = -12.
∫₀^π e^x sin x dx = e^π2 + 12 = e^π + 12.
Answer: option (a).

Answer: (b) e² + 14

Method. integration by parts with u = ln x, dv = x dx

∫ x ln x dx = x²2 ln x - ∫ x²2·1x dx = x²2 ln x − x²4.
At x = e: e²2 − e²4 = e²4.
At x = 1: 0 − 14 = -14.
∫₁^e x ln x dx = e²4 + 14 = e² + 14.
Answer: option (b).

Answer: (a) -1, 1

Method. evaluate both integrals and solve b⁴ = a⁴, b³ - a³ = 2

∫_a^b x³ dx = b⁴ - a⁴4 = 0, so b⁴ = a⁴, i.e. b = a or b = -a.
∫_a^b x² dx = b³ - a³3 = 23, so b³ - a³ = 2.
b = a would give 0 ≠ 2, so b = -a: -a³ - a³ = 2, a³ = -1, a = -1, b = 1.
Check: ∫_-1¹ x³ dx = 0 and ∫_-1¹ x² dx = 23.
Answer: a = -1, b = 1, option (a).

Shortcut. Test the options: for (a), x³ is odd so ∫_-1¹ x³ = 0 and ∫_-1¹ x² = 23; (b), (c) give 0 for the second integral and (d) gives -163.

Answer: (a) 1/⁠110

Method. property ∫₀¹ h(x)dx = ∫₀¹ h(1 − x)dx, then power rule

Replace x by 1 − x: ∫₀¹ x(1 − x)⁹ dx = ∫₀¹ (1 − x)x⁹ dx.
= ∫₀¹ (x⁹ - x¹⁰) dx = 110 − 111.
= 11 − 10110 = 1110.
Answer: option (a).

Answer: (a) 8

Method. write 1 + sin t as a perfect square (sint2 + cost2)²

Put x = 2t, dx = 2dt; t runs from 0 to π: I = 2∫₀^π ⁠√(1⁠ + sin t) dt.
1 + sin t = sin²t2 + cos²t2 + 2sint2cost2 = (sint2 + cost2)².
For 0 ≤ t ≤ π, both sint2 and cost2 are ≥ 0, so ⁠√(1⁠ + sin t) = sint2 + cost2.
∫₀^π (sint2 + cost2)dt = [-2cost2 + 2sint2]₀^π = (0 + 2) − (-2 + 0) = 4.
I = 2 × 4 = 8.
Answer: 8, option (a).

Answer: (b) 1

Method. half-angle identity 1 + cos θ = 2cos²(θ/⁠2)

1 + cos θ = 2cos²θ2, so 11 + cos θ = 12sec²θ2.
∫ 12sec²θ2 dθ = tanθ2.
∫₀^π/⁠2 dθ1 + cos θ = tanπ4 − tan 0 = 1 − 0 = 1.
Answer: 1, option (b).

Answer: (c) 12

Method. x - [x] = x − n on [n, n + 1)

For n ≤ x < n + 1, [x] = n, so x - [x] = x − n.
∫_nⁿ⁺¹ (x − n) dx = [(x − n)²2]_nⁿ⁺¹ = 12.

Answer: (b) 2(⁠√3⁠ - ⁠√2)⁠

Method. [x²] is constant on [⁠√2,⁠ ⁠√3)⁠

For ⁠√2⁠ ≤ x < ⁠√3:⁠ 2 ≤ x² < 3, so f(x) = [x²] = 2.
∫_⁠√2^√3⁠ 2 dx = 2(⁠√3⁠ - ⁠√2).⁠

Answer: (a) 6 - ⁠√3⁠ − 2⁠√2⁠

Method. split at ⁠√3,⁠ where [x²] jumps from 2 to 3

⁠√2⁠ ≤ x < ⁠√3:⁠ 2 ≤ x² < 3, f(x) = 2.
⁠√3⁠ ≤ x < 2: 3 ≤ x² < 4, f(x) = 3.
∫_⁠√2²⁠ f(x) dx = 2(⁠√3⁠ - ⁠√2)⁠ + 3(2 - ⁠√3)⁠ = 2⁠√3⁠ − 2⁠√2⁠ + 6 − 3⁠√3⁠ = 6 - ⁠√3⁠ − 2⁠√2.⁠

Answer: (d) 10/⁠3

Method. sign of x² - x − 2 = (x − 2)(x + 1) on [0, 2]

x² - x − 2 = (x − 2)(x + 1) ≤ 0 for -1 ≤ x ≤ 2, so on [0, 2] f(x) = 2 + x − x².
∫₀² (2 + x − x²) dx = [2x + x²2 − x³3]₀² = 4 + 2 − 83 = 103.

Answer: (b) 3

Method. split at the root x = 2 where x² - x − 2 changes sign

On [1, 2]: f(x) = 2 + x − x²; on [2, 3]: f(x) = x² - x − 2.
∫₁² (2 + x − x²) dx = [2x + x²2 − x³3]₁² = 103 − 136 = 76.
∫₂³ (x² - x − 2) dx = [x³3 − x²2 − 2x]₂³ = (9 − 92 − 6) − (83 − 2 − 4) = -32 + 103 = 116.
Total = 76 + 116 = 3.

Answer: (d) p = q

Method. |f(x)| = f(x) when f(x) > 0

e^-x > 0 for every real x, so |f(x)| = |e^-x| = e^-x = f(x).
Then q = ∫_a^b |f(x)| dx = ∫_a^b f(x) dx = p, i.e. p = q.
(Here p = e^-a - e^-b ≠ 0 for a ≠ b, so p = 2q, p = -q and 4p = q all fail.)

Answer: (a) -32

Method. on [-1, 0) the greatest integer [x] = -1

For -1 ≤ x < 0: [x] = -1 and x − 1 < 0, so |x − 1| = 1 − x and h(x) = -(1 − x) = x − 1.
∫_-1⁰ (x − 1) dx = [x²2 − x]_-1⁰ = 0 − (12 + 1) = -32.

Answer: (d) 12

Method. split [0, 2] at the integer 1

0 ≤ x < 1: [x] = 0, so h(x) = 0.
1 ≤ x < 2: [x] = 1 and x − 1 ≥ 0, so h(x) = x − 1.
∫₀² h(x) dx = 0 + ∫₁² (x − 1) dx = [(x − 1)²2]₁² = 12.

Answer: (a) a + b

Method. split at 0: |x|x = -1 for x < 0 and 1 for x > 0

For a < 0 < b: l = ∫_a⁰ (-1) dx + ∫₀^b 1 dx.
= -(0 − a) + (b − 0) = a + b.

Answer: (b) a − b

Method. |x|x = -1 throughout when x < 0

For a < b < 0 the whole interval is negative, so |x|x = -1.
l = ∫_a^b (-1) dx = -(b − a) = a − b.

Answer: (b) -1

Method. for x < 0, |x| = -x, so x|x| = -1

On [-2, -1], x < 0, so |x| = -x and x|x| = -1.
∫_-2^-1 (-1) dx = -(-1 − (-2)) = -1.
Answer: option (b).

Answer: (d) greater than or equal to k

Method. |∫ f| ≤ ∫ |f|, together with k ≤ |k|

Since -|f(x)| ≤ f(x) ≤ |f(x)|, integrating over [-2, 0] gives ∫_-2⁰ f(x) dx ≤ ∫_-2⁰ |f(x)| dx.
So ∫_-2⁰ |f(x)| dx ≥ k.
Equality is possible (for example when f(x) ≥ 0 on [-2, 0]), so 'greater than k' is not always true; the correct statement is 'greater than or equal to k'.
Answer: option (d).

Answer: (b) 2(⁠√2⁠ − 1)

Method. split at x = π4, where sin x = cos x

On [0, π4], cos x ≥ sin x; on [π4, π2], sin x ≥ cos x.
∫₀^π/⁠4 (cos x − sin x) dx = [sin x + cos x]₀^π/⁠4 = ⁠√2⁠ − 1.
∫_π/⁠4^π/⁠2 (sin x − cos x) dx = [-cos x − sin x]_π/⁠4^π/⁠2 = -1 + ⁠√2⁠ = ⁠√2⁠ − 1.
Total = 2(⁠√2⁠ − 1).
Answer: option (b).

Answer: (b) a − b

Method. [x] + [-x] = -1 when x is not an integer, 0 when x is an integer

If x is an integer, [x] + [-x] = x − x = 0.
If x is not an integer, write x = n + f with 0 < f < 1: [x] = n and [-x] = [-n − f] = -n − 1, so [x] + [-x] = -1.
On [a, b] the integers are finitely many points, which do not change the integral.
∫_a^b [x] dx + ∫_a^b [-x] dx = ∫_a^b ([x] + [-x]) dx = ∫_a^b (-1) dx = -(b − a) = a − b.
Answer: option (b).

Answer: (d) 9

Method. split |x − 5| at x = 5

For 2 ≤ x ≤ 5, |x − 5| = 5 − x; for 5 ≤ x ≤ 8, |x − 5| = x − 5.
∫₂⁵ (5 − x) dx = [5x − x²2]₂⁵ = 252 − 8 = 92.
∫₅⁸ (x − 5) dx = [x²2 − 5x]₅⁸ = -8 − (-252) = 92.
Total = 92 + 92 = 9.
Answer: option (d).

Shortcut. The graph of |x − 5| on [2, 8] is two triangles of base 3 and height 3: area = 2 × 12 × 3 × 3 = 9.

Answer: (a) ⁠√2⁠ − 1

Method. split the range where [x²] is constant

For 0 ≤ x < 1: 0 ≤ x² < 1, so [x²] = 0.
For 1 ≤ x < ⁠√2:⁠ 1 ≤ x² < 2, so [x²] = 1.
∫₀^⁠√2⁠ [x²] dx = ∫₀¹ 0 dx + ∫₁^⁠√2⁠ 1 dx = ⁠√2⁠ − 1.
Answer: option (a).

Answer: (b) 52

Method. substitution u = ln x, then split |u| at u = 0

Put u = ln x, du = dxx; x = e^-1 gives u = -1, x = e² gives u = 2.
The integral becomes ∫_-1² |u| du.
= ∫_-1⁰ (-u)du + ∫₀² u du = 12 + 42 = 12 + 2 = 52.
Answer: 52, option (b).

Answer: (a) π4

Method. property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx

Let J = ∫₀^π/⁠2 a sin x + b cos xsin x + cos x dx, so k = Ja + b.
Put x → π2 − x: J = ∫₀^π/⁠2 a cos x + b sin xsin x + cos x dx.
Adding: 2J = ∫₀^π/⁠2 (a + b)(sin x + cos x)sin x + cos x dx = (a + b)π2, so J = (a + b)π4.
k = Ja + b = π4.

Answer: (c) k(a + b)

Method. property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx links the integral to k

Put x → π2 − x in the required integral: ∫₀^π/⁠2 a cos x + b sin xsin x + cos x dx = ∫₀^π/⁠2 a sin x + b cos xcos x + sin x dx.
By the definition of k, ∫₀^π/⁠2 a sin x + b cos xsin x + cos x dx = k(a + b).
So the required integral = k(a + b) (= (a + b)π4).

No correct option. This question is defective as printed in UPSC's paper.

Method. property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx (King's rule), checked against where g vanishes

g(x) = sin x + sin(4 − x) = 2 sin 2 cos(x − 2), and g(4 − x) = g(x).
Formally, with x → 4 − x: ∫₀⁴ f(x)g(x) dx = ∫₀⁴ f(4 − x)g(x) dx, and adding gives 2∫₀⁴ f(x)g(x) dx = ∫₀⁴ 1 dx = 4, i.e. the value 2.
But g(x) = 0 where cos(x − 2) = 0, i.e. at x = 2 − π2 ≈ 0.429 and x = 2 + π2 ≈ 3.571, both inside [0, 4], while sin x ≠ 0 there (sin(2 ∓ π2) = ∓cos 2 ≈ ±0.416).
Near each such point f(x)g(x) behaves like cx − x₀ with c ≠ 0, so the integral diverges; the value 2 is only a principal value.

g(x) = sin x + sin(4 − x) = 2 sin 2 cos(x − 2) vanishes at x = 2 ± π/⁠2 inside [0, 4] where sin x ≠ 0, so the integrand has non-integrable poles and the integral does not exist; the evident intended King's-rule reading gives 2, option (c).

No correct option. This question is defective as printed in UPSC's paper.

Method. substitution x → 4 − x reduces it to the previous item

With u = 4 − x (dx = -du, limits 4 → 0): ∫₀⁴ f(4 − x)g(4 − x) dx = ∫₀⁴ f(u)g(u) du, the integral of the previous item.
Formally that equals 2 (King's rule), but g(u) = 2 sin 2 cos(u − 2) vanishes at u = 2 ± π2 inside [0, 4] while sin u ≠ 0 there, so the integrand has simple poles and the integral diverges.

The integral equals ∫₀⁴ f(x)/⁠g(x) dx, whose integrand has non-integrable poles at x = 2 ± π/⁠2 (where g = 2 sin 2 cos(x − 2) = 0 but sin x ≠ 0), so it does not exist; the evident intended reading gives 2, option (c).

Answer: (a) π4

Method. property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx

Let I = ∫₀^π/⁠2 a + sin x2a + sin x + cos x dx.
Put x → π2 − x: sin ↔ cos, the denominator is unchanged, so I = ∫₀^π/⁠2 a + cos x2a + sin x + cos x dx.
Adding: 2I = ∫₀^π/⁠2 2a + sin x + cos x2a + sin x + cos x dx = ∫₀^π/⁠2 1 dx = π2.
Optionally append to the working: "(valid whenever 2a + sin x + cos x ≠ 0 on [0, π/⁠2], i.e. a ∉ [-1/⁠√2,⁠ -1/⁠2]; the question intends such a.)"

No correct option. This question is defective as printed in UPSC's paper.

Method. property ∫₀^a x f(x) dx = a2∫₀^a f(x) dx when f(a − x) = f(x), then the ratio I₁/⁠I₂

I₂ = 2∫₀^π/⁠2 dx1 + sin² x = 2∫₀^π/⁠2 sec² x dx1 + 2tan² x = 2 · 1⁠√2[tan^-1(√2⁠ tan x)]₀^π/⁠2 = π⁠√2.⁠
In I₁, put x → π − x: I₁ = ∫₀^π π − x1 + cos² x dx, so 2I₁ = π∫₀^π dx1 + cos² x = π · π⁠√2⁠ (the same value as I₂, since cos² x and sin² x run through the same values over a full period π).
So I₁ = π²2⁠√2⁠ = π2 I₂.
I₁ + I₂I₁ - I₂ = π/⁠2 + 1π/⁠2 − 1 = π + 2π − 2 ≈ 4.504.
The options are 1, π ≈ 3.142, π² ≈ 9.870 and π + 1π − 1 ≈ 1.934: none equals π + 2π − 2.

With I₁ = π²/⁠(2⁠√2)⁠ and I₂ = π/⁠√2⁠ as printed, the ratio is (π + 2)/⁠(π − 2), which is none of the options; option (d) (π + 1)/⁠(π − 1) would follow if I₁ = πI₂, e.g. if I₂ were taken over [0, π/⁠2] (value π/⁠(2⁠√2)).⁠

Answer: (d) π⁴

Method. property ∫₀^a x f(x) dx = a2∫₀^a f(x) dx with f(π − x) = f(x)

Put x → π − x: I₁ = ∫₀^π π − x1 + cos² x dx; adding, 2I₁ = π∫₀^π dx1 + cos² x.
∫₀^π dx1 + cos² x = 2∫₀^π/⁠2 sec² x dxsec² x + 1 = 2∫₀^π/⁠2 sec² x dx2 + tan² x = 2 · 1⁠√2[tan^-1tan⁠ x⁠√2]₀^π/⁠2⁠ = π⁠√2.⁠
So I₁ = π²2⁠√2,⁠ I₁² = π⁴8.
8I₁² = π⁴.

Answer: (a) π⁠√2⁠

Method. substitution t = tan x after writing the integral over [0, π/⁠2]

sin²(π − x) = sin² x, so I₂ = 2∫₀^π/⁠2 dx1 + sin² x.
Divide by cos² x: sec² xsec² x + tan² x = sec² x1 + 2tan² x.
Put t = tan x (0 → ∞): I₂ = 2∫₀^∞ dt1 + 2t² = 2 · 1⁠√2[tan^-1(√2⁠ t)]₀^∞ = 2⁠√2⁠ · π2.
I₂ = π⁠√2.⁠

Answer: (b) 0

Method. ln(1 − xx) changes sign under x → 1 − x

1x − 1 = 1 − xx, so I = ∫₀¹ ln(1 − xx) dx (an improper integral that converges, as ∫₀¹ ln x dx = -1 does).
Replace x by 1 − x: I = ∫₀¹ ln(x1 − x) dx = -I.
So 2I = 0, I = 0.
Answer: option (b).

Answer: (a) 0

Method. substitution x = 2t, then ∫₀^π/⁠2 ln(tan t)dt = 0 by t → π2 − t

Put x = 2t, dx = 2dt; t runs from 0 to π2: I = 2∫₀^π/⁠2 ln(tan t) dt.
Let J = ∫₀^π/⁠2 ln(tan t) dt. Replace t by π2 − t: J = ∫₀^π/⁠2 ln(cot t) dt = -∫₀^π/⁠2 ln(tan t) dt = -J.
So 2J = 0, J = 0, and I = 0.
Answer: option (a).

Shortcut. Directly: replacing x by π − x turns tanx2 into cotx2, so I = -I and I = 0.

Answer: (d) a2

Method. property ∫₀^a h(x)dx = ∫₀^a h(a − x)dx, then add

Let I = ∫₀^a f(a − x)f(x) + f(a − x) dx (with f(x) + f(a − x) ≠ 0 on [0, a]).
Replace x by a − x: I = ∫₀^a f(x)f(a − x) + f(x) dx.
Adding: 2I = ∫₀^a f(a − x) + f(x)f(x) + f(a − x) dx = ∫₀^a 1 dx = a.
I = a2.
Answer: option (d).

Answer: (c) π

Method. property ∫₀^π x h(sin x)dx = π2∫₀^π h(sin x)dx

Replace x by π − x (sin(π − x) = sin x): I₁ = ∫₀^π (π − x)dx1 + sin x.
Adding: 2I₁ = π∫₀^π dx1 + sin x.
11 + sin x = 1 − sin xcos² x = sec² x − sec x tan x, with antiderivative tan x − sec x.
∫₀^π dx1 + sin x = (tan π − sec π) − (tan 0 − sec 0) = (0 + 1) − (0 − 1) = 2 (the integrand is continuous on [0, π] and tan x − sec x = sin x − 1cos x → 0 at x = π2, so the antiderivative is continuous there).
2I₁ = 2π, so I₁ = π.
Answer: option (c).

Answer: (a) 2π

Method. sin(π + x) = -sin x, then x → π − x shows I₂ = I₁

sin(π + x) = -sin x, so I₂ = ∫₀^π (π − x)dx1 + sin x.
Replace x by π − x (sin(π − x) = sin x): I₂ = ∫₀^π x dx1 + sin x = I₁.
From the previous item, I₁ = π.
I₁ + I₂ = 2π.
Answer: option (a).

Answer: (d) π⁠√2⁠

Method. combine into sin x + cos x⁠√(sin⁠ x cos x) and substitute u = sin x − cos x

⁠√(tan⁠ x) + ⁠√(cot⁠ x) = sin x + cos x⁠√(sin⁠ x cos x) on (0, π4].
(sin x − cos x)² = 1 − 2sin x cos x, so sin x cos x = 1 − (sin x − cos x)²2.
So the sum of the integrals is I = ⁠√2⁠ ∫₀^π/⁠4 (sin x + cos x)dx⁠√(1⁠ −⁠ (sin x −⁠ cos x)²).
Put u = sin x − cos x, du = (cos x + sin x)dx; x = 0 gives u = -1, x = π4 gives u = 0.
I = ⁠√2⁠ ∫_-1⁰ du⁠√(1⁠ −⁠ u²) = ⁠√2⁠ [sin^-1 u]_-1⁰ = ⁠√2⁠ (0 + π2) = π⁠√2.⁠
Answer: option (d).

Answer: (b) ∫₀^a f(x)dx

Method. property ∫₀^a h(x)dx = ∫₀^a h(a − x)dx

Let I = ∫₀^a f(x)g(x)dx.
Replace x by a − x: I = ∫₀^a f(a − x)g(a − x)dx = ∫₀^a f(x)[2 − g(x)]dx, using f(a − x) = f(x) and g(a − x) = 2 − g(x).
So I = 2∫₀^a f(x)dx - ∫₀^a f(x)g(x)dx = 2∫₀^a f(x)dx − I.
2I = 2∫₀^a f(x)dx, so I = ∫₀^a f(x)dx.
Answer: option (b).

Answer: (a) 0

Method. f(x + y) = f(x) + f(y) forces f(0) = 0 and f(-x) = -f(x), so f is odd

Put x = y = 0: f(0) = 2f(0), so f(0) = 0.
Put y = -x: f(0) = f(x) + f(-x), so f(-x) = -f(x); f is odd.
The integral of an odd function over [-1, 1] is 0.

Answer: (b) 0

Method. integral of an odd function over [-a, a] is 0

Let f(x) = (3 sin x − sin 3x) cos² x.
f(-x) = (-3 sin x + sin 3x) cos² x = -f(x), so f is odd.
Hence ∫_-1¹ f(x) dx = 0.

Answer: (b) 3π4

Method. sin⁴ x + cos⁴ x = 1 − 12sin² 2x

sin⁴ x + cos⁴ x = (sin² x + cos² x)² - 2sin² x cos² x = 1 − 12sin² 2x = 1 − 1 − cos 4x4 = 34 + cos 4x4.
∫₀^π (34 + cos 4x4) dx = 3π4 + 116[sin 4x]₀^π = 3π4.

Answer: (c) 3π2

Method. ∫_-a^a g(x)1 + b^x dx = ∫₀^a g(x) dx for even g

Let g(x) = sin⁴ x + cos⁴ x (even). Put x → -x: I = ∫_-2π^2π g(x)1 + 3^-x dx = ∫_-2π^2π 3^x g(x)3^x + 1 dx.
Adding the two forms: 2I = ∫_-2π^2π g(x) dx = 2∫₀^2π g(x) dx, so I = ∫₀^2π g(x) dx.
g has period π/⁠2, so ∫₀^2π g = 2∫₀^π g = 2 · 3π4 (previous item).
I = 3π2.

Answer: (a) e² - 1e

Method. the e^{cos x} sin x part is odd and integrates to 0; the other part is ∫ e^tdt with t = sin x

e^cos(-x) sin(-x) = -e^{cos x} sin x, so e^{cos x} sin x is odd and ∫_-π/⁠2^π/⁠2 e^{cos x} sin x dx = 0.
For the second part put t = sin x, dt = cos x dx; x = -π2 gives t = -1, x = π2 gives t = 1.
∫_-1¹ e^t dt = e − e^-1 = e² - 1e.
Total = 0 + e² - 1e.
Answer: option (a).

Answer: (a) f(x)

Method. substitute x → -x in ∫₀^a f(-x)dx to turn it into ∫_-a⁰ f(x)dx

∫₀^a [f(x) + f(-x)] dx = ∫₀^a f(x) dx + ∫₀^a f(-x) dx.
In the second, put x = -t, dx = -dt: ∫₀^a f(-x) dx = ∫₀^-a f(t)(-dt) = ∫_-a⁰ f(t) dt.
So the left side = ∫_-a⁰ f(x) dx + ∫₀^a f(x) dx = ∫_-a^a f(x) dx.
After 'Hence g(x) = f(x) satisfies the equation for every f.' add: 'The equation fixes only the integral of g, not g itself (g(x) = f(-x) would also do), so the question is read as asking which listed option works for every f; only (a) does.'
Option (b) would give ∫_-a^a [f(x) + f(-x)] dx = 2∫₀^a [f(x) + f(-x)] dx, twice the left side, and (c) gives its negative.
Answer: option (a).

Answer: (c) 0

Method. integral of an odd function over [-a, a] is 0

Let h(x) = sin x − tan x. Then h(-x) = -sin x + tan x = -h(x), so h is odd.
h is continuous on [-π4, π4] (tan x is defined there).
So ∫_-π/⁠4^π/⁠4 (sin x − tan x) dx = 0.
Answer: option (c).

Answer: (d) 200

Method. periodicity: |sin x| has period π, so the integral over any π-long interval is the same

For a periodic function of period π, ∫_a^{a + π} |sin x| dx = ∫₀^π |sin x| dx = [-cos x]₀^π = 2, whatever a is.
[a, a + 100π] is 100 such intervals: φ(a) = 100 × 2 = 200.

Answer: (a) 0

Method. φ(a) is constant (or Leibniz rule)

From the previous item φ(a) = 200 for every a, so φ'(a) = 0.
Check by the Leibniz rule: φ'(a) = |sin(a + 100π)| · 1 − |sin a| · 1 = |sin a| − |sin a| = 0.

Answer: (d) 16

Method. periodicity: |sin x| has period π

∫₀^π |sin x| dx = ∫₀^π sin x dx = [-cos x]₀^π = 2.
[0, 8π] holds 8 periods, so ∫₀^8π |sin x| dx = 8 × 2 = 16.

Answer: (d) 40k

Method. sin⁴x + cos⁴x has period π2; ∫ over n periods = n × ∫ over one period

sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x cos²x = 1 − 12sin² 2x = 34 + 14cos 4x.
cos 4x has period 2π4 = π2, so the integrand has period π2.
[0, 20π] is made of 20ππ/⁠2 = 40 such periods, each contributing ∫₀^π/⁠2 (sin⁴x + cos⁴x)dx = k.
∫₀^20π (sin⁴x + cos⁴x)dx = 40k.
Answer: option (d).