Integrals Class 12 NCERT Solutions Chapter 7 Misc

Concept used. Factor and apply partial fractions. x - x³ = x(1 - x²) = x(1 - x)(1 + x).

1. Decomposition: 1x(1 - x)(1 + x) = Ax + B1 - x + C1 + x.
2. Multiply by denominator: 1 = A(1 - x)(1 + x) + B x(1 + x) + C x(1 - x). At x = 0: 1 = A · 1· 1 = A ⇒ A = 1. At x = 1: 1 = B· 1· 2 = 2 B ⇒ B = 1/2. At x = -1: 1 = C(-1)(2) = -2 C ⇒ C = -1/2.
3. Integrate: I = ∫ [1x + 1/21 - x + -1/21 + x] dx = log|x| - 12log|1 - x| - 12log|1 + x| + C.
4. Combine logs: I = log|x| - 12log|1 - x²| + C = 12log|x²1 - x²| + C.

12log|x²1 - x²| + C

Concept used. Rationalise by multiplying by the conjugate over itself.

1. Multiply by √x+a - √x+b√x+a - √x+b: 1√x+a + √x+b· √x+a - √x+b√x+a - √x+b = √x+a - √x+b(x + a) - (x + b) = √x+a - √x+ba - b.
2. Integrate each piece: ∫ √x + a dx = 2(x + a)^3/23, similarly for √x + b.
3. Combine: I = 1a - b[2(x + a)^3/23 - 2(x + b)^3/23] + C = 23(a - b)[(x + a)^3/2 - (x + b)^3/2] + C.

23(a - b)[(x + a)^3/2 - (x + b)^3/2] + C

Concept used. Substitution per the hint. Let x = a/t, so dx = -a/t² dt. Then a x - x² = a(a/t) - a²/t² = a² t - a²t² = a²(t - 1)t², so √a x - x² = a|t|√t - 1 (assume a > 0, t > 0).

1. x√a x - x² = at· a√t - 1t = a²√t - 1t².
2. dxx√a x - x² = -a/t² dta²√t - 1/t² = -dta√t - 1.
3. Integrate: I = -1a∫ dt√t - 1 = -1a· 2√t - 1 + C = -2√t - 1a + C.
4. Back-substitute t = a/x: I = -2a√ax - 1 + C = -2a√a - xx + C.

-2a√a - xx + C

Concept used. Factor x⁴ from inside the bracket. (x⁴ + 1)^3/4 = x³ (1 + 1/x⁴)^3/4 (for x > 0). So the integrand is 1x² · x³ (1 + 1/x⁴)^3/4 = 1x⁵ (1 + 1/x⁴)^3/4. Substitute t = 1 + 1/x⁴, dt = -4/x⁵ dx, so dx/x⁵ = -dt/4.

1. Transform: I = ∫ -dt/4t^3/4 = -14∫ t^-3/4 dt = -14· 4 t^1/4 + C = -t^1/4 + C.
2. Back-substitute: I = -(1 + 1/x⁴)^1/4 + C = -(x⁴ + 1)^1/4x + C.

-(x⁴ + 1)^1/4x + C

Concept used. Pick t = x^1/6 so x = t⁶ and dx = 6 t⁵ dt. Then x^1/2 = t³ and x^1/3 = t², eliminating fractional exponents.

1. Sub: dxx^1/2 + x^1/3 = 6 t⁵ dtt³ + t² = 6 t⁵ dtt²(t + 1) = 6 t³ dtt + 1.
2. Divide t³ by t + 1: t³ = (t + 1)(t² - t + 1) - 1, so t³t + 1 = t² - t + 1 - 1t + 1.
3. Integrate: I = 6∫ (t² - t + 1 - 1t + 1) dt = 6[t³3 - t²2 + t - log|t + 1|] + C.
4. Restore t = x^1/6: I = 2 x^1/2 - 3 x^1/3 + 6 x^1/6 - 6log|x^1/6 + 1| + C.

2x - 3 x^1/3 + 6 x^1/6 - 6log|x^1/6 + 1| + C

Concept used. Partial fractions with a linear and an irreducible quadratic factor. Write 5 x(x + 1)(x² + 9) = Ax + 1 + B x + Cx² + 9.

1. Multiply: 5 x = A(x² + 9) + (B x + C)(x + 1) = (A + B)x² + (B + C)x + (9 A + C).
2. Match coefficients: A + B = 0; B + C = 5; 9 A + C = 0.
3. Solve: from the first, B = -A; substituting into the second, C = 5 + A. Third: 9 A + 5 + A = 0 ⇒ 10 A = -5 ⇒ A = -1/2. Then B = 1/2, C = 9/2.
4. Decomposition: 5 x(x + 1)(x² + 9) = -1/2x + 1 + (1/2) x + 9/2x² + 9.
5. Integrate piece-wise: ∫ -1/2x + 1 dx = -12log|x + 1|; ∫ (1/2) xx² + 9 dx = 14log(x² + 9); ∫ 9/2x² + 9 dx = 92· 13tan^-1(x/3) = 32tan^-1(x/3).
6. Combine: I = -12log|x + 1| + 14log(x² + 9) + 32tan^-1x3 + C.

-12log|x + 1| + 14log(x² + 9) + 32tan^-1x3 + C

Concept used. Write x = (x - a) + a and expand sin x = sin((x - a) + a) using the sum formula.

1. Expand: sin x = sin(x - a)cos a + cos(x - a)sin a.
2. Divide by sin(x - a): sin xsin(x - a) = cos a + sin a · cot(x - a).
3. Integrate: ∫ cos a dx = xcos a; ∫ cot(x - a) dx = log|sin(x - a)|.
4. Combine: I = xcos a + sin a · log|sin(x - a)| + C.

xcos a + sin a · log|sin(x - a)| + C

Concept used. e^{nlog x} = xⁿ. Simplify before integrating.

1. Convert: numerator = x⁵ - x⁴ = x⁴(x - 1); denominator = x³ - x² = x²(x - 1).
2. Cancel (x - 1): ratio = x⁴/x² = x².
3. Integrate: I = ∫ x² dx = x³3 + C.

x³/3 + C

Concept used. Substitution t = sin x converts to ∫ dt√4 - t² = sin^-1(t/2) + C.

1. t = sin x ⇒ dt = cos x dx.
2. Transform: I = ∫ dt√4 - t² = sin^-1t2 + C.
3. Restore: I = sin^-1sin x2 + C.

sin^-1sin x2 + C

Concept used. Factor as a difference of squares (twice) and use sin² + cos² = 1; the denominator equals sin⁴ + cos⁴.

1. Note sin⁴ + cos⁴ = (sin² + cos²)² - 2sin²cos² = 1 - 2sin²cos². So the denominator equals sin⁴ x + cos⁴ x.
2. Difference of squares: sin⁸ - cos⁸ = (sin⁴ - cos⁴)(sin⁴ + cos⁴).
3. Cancel the common factor sin⁴ + cos⁴ (positive): Integrand = sin⁴ x - cos⁴ x = (sin² x - cos² x)(sin² x + cos² x) = -cos 2 x. (Used sin² - cos² = -cos 2 x and sin² + cos² = 1.)
4. Integrate: ∫ -cos 2 x dx = -sin 2 x2 + C.

-sin 2 x2 + C

Concept used. Multiply and divide by sin(a - b), then use sin((x + a) - (x + b)) = sin(a - b) to break up the integrand.

1. Expand sin(a - b) = sin((x + a) - (x + b)) = sin(x + a)cos(x + b) - cos(x + a)sin(x + b).
2. Divide by cos(x+a)cos(x+b): sin(a - b)cos(x + a)cos(x + b) = tan(x + a) - tan(x + b).
3. So 1cos(x + a)cos(x + b) = 1sin(a - b)[tan(x + a) - tan(x + b)].
4. Integrate: ∫ tan u du = log|sec u|. So I = 1sin(a - b)[log|sec(x + a)| - log|sec(x + b)|] + C. Combining the logs: I = 1sin(a - b)log|cos(x + b)cos(x + a)| + C.

1sin(a - b)log|cos(x + b)cos(x + a)| + C

Concept used. Substitute t = x⁴, dt = 4 x³ dx.

1. Transform: 1 - x⁸ = 1 - (x⁴)² = 1 - t². And x³ dx = dt/4.
2. Integrate: I = ∫ dt/4√1 - t² = 14sin^-1 t + C.
3. Restore: I = 14sin^-1(x⁴) + C.

14sin^-1(x⁴) + C

Concept used. Substitute t = e^x, dt = e^x dx. Then partial fractions.

1. Transform: e^x dx(1 + e^x)(2 + e^x) = dt(1 + t)(2 + t).
2. Partial fractions: 1(1+t)(2+t) = 11+t - 12+t.
3. Integrate: I = ∫ (11 + t - 12 + t) dt = log|1 + t| - log|2 + t| + C = log|1 + t2 + t| + C.
4. Restore t = e^x: I = log|1 + e^x2 + e^x| + C.

log|1 + e^x2 + e^x| + C

Concept used. Partial fractions on two irreducible quadratics.

1. Write 1(x² + 1)(x² + 4) = Ax² + 1 + Bx² + 4.
2. Multiply: 1 = A(x² + 4) + B(x² + 1) = (A + B)x² + (4 A + B).
3. Match: A + B = 0 and 4 A + B = 1 ⇒ 3 A = 1 ⇒ A = 1/3, B = -1/3.
4. Integrate: I = 13∫ dxx² + 1 - 13∫ dxx² + 4 = 13tan^-1 x - 13· 12tan^-1(x/2) + C.
5. Tidy: I = 13tan^-1 x - 16tan^-1x2 + C.

13tan^-1 x - 16tan^-1x2 + C

Concept used. e^x = sin x. So the integrand simplifies to sin x · cos³ x, a textbook t = cos x substitution.

1. Simplify: integrand = sin x cos³ x.
2. Substitute t = cos x, dt = -sin x dx, so sin x dx = -dt.
3. Transform: I = ∫ t³ · (-dt) = -t⁴4 + C.
4. Restore: I = -cos⁴ x4 + C.

-cos⁴ x4 + C

Concept used. e^{3log x} = x³. So the integrand is x³x⁴ + 1.

1. Substitute t = x⁴ + 1, dt = 4 x³ dx, so x³ dx = dt/4.
2. Transform: I = ∫ dt/4t = 14log|t| + C = 14log(x⁴ + 1) + C.

14log(x⁴ + 1) + C

Concept used. Generalised power rule with a substitution. Let u = f(ax + b), then du = f'(ax + b)· a dx, so f'(ax + b) dx = du/a.

1. Transform: f'(ax + b) [f(ax + b)]ⁿ dx = uⁿ · dua.
2. Integrate: 1a∫ uⁿ du = uⁿ⁺¹a(n+1) + C.
3. Restore: I = [f(ax + b)]^{n + 1}a(n + 1) + C.

[f(ax + b)]^{n + 1}a(n + 1) + C

Concept used. Factor and use sin(x + α) = sin xcosα + cos xsinα, then substitute t = cot x + cosα/sin x or, more compactly, t = cosα + cot x sinα — a manipulation that turns the integrand into -d t / (2t sinα). The standard NCERT solution uses t = sin(x + α)sin x.

1. Note sin(x + α)sin x = cosα + cot xsinα. So let t = cosα + cot x sinα. Then dtdx = -cosec² x sinα, dt = -sinαsin² x dx.
2. Rewrite the integrand. sin³ x sin(x + α) = sin⁴ x · sin(x + α)sin x = sin⁴ x · t. So √sin³ x sin(x + α) = sin² x t.
3. Transform: dxsin² x t = 1t· dxsin² x = -1sinα· dtt.
4. Integrate: I = -1sinα∫ dtt = -2tsinα + C.
5. Restore: I = -2sinα√cosα + cot xsinα + C = -2sinα√sin(x + α)sin x + C.

-2sinα√sin(x + α)sin x + C

Concept used. Trig substitution x = cosθ removes the radical via the half-angle identity √(1 - cosθ)/(1 + cosθ) = tan(θ/2).

1. Substitute x = cosθ, dx = -sinθ dθ, θ = cos^-1 x.
2. Integrand: √(1 - cosθ)/(1 + cosθ) = tan(θ/2).
3. Transform: I = ∫ tan(θ/2)· (-sinθ) dθ. Use sinθ = 2sin(θ/2)cos(θ/2): tan(θ/2)sinθ = sin(θ/2)cos(θ/2)· 2sin(θ/2)cos(θ/2) = 2sin²(θ/2) = 1 - cosθ.
4. So I = -∫ (1 - cosθ) dθ = -θ + sinθ + C.
5. Restore: θ = cos^-1 x, sinθ = √1 - x²: I = √1 - x² - cos^-1 x + C.

√1 - x² - cos^-1 x + C

Concept used. Use 1 + cos 2 x = 2cos² x and sin 2 x = 2sin xcos x, then recognise the e^x[f + f'] template.

1. Simplify: 2 + sin 2 x1 + cos 2 x = 2 + 2sin xcos x2cos² x = 1cos² x + sin xcos x = sec² x + tan x.
2. Set f(x) = tan x. Then f'(x) = sec² x, so f + f' = tan x + sec² x.
3. Apply template: I = e^x f(x) + C = e^x tan x + C.

e^x tan x + C

Concept used. Partial fractions with a repeated linear factor. x² + x + 1(x + 1)²(x + 2) = Ax + 1 + B(x + 1)² + Cx + 2.

1. Multiply: x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)².
2. x = -1: 1 - 1 + 1 = 1 = B(1) ⇒ B = 1.
3. x = -2: 4 - 2 + 1 = 3 = C(1) ⇒ C = 3.
4. Expand and match x² coefficient: A + C = 1 ⇒ A = 1 - 3 = -2.
5. Decomposition: -2x + 1 + 1(x + 1)² + 3x + 2.
6. Integrate: I = -2log|x + 1| - 1x + 1 + 3log|x + 2| + C.

3log|x + 2| - 2log|x + 1| - 1x + 1 + C

Concept used. Substitute x = cosθ. Use the identity √(1 - cosθ)/(1 + cosθ) = tan(θ/2).

1. Let x = cosθ ⇒ dx = -sinθ dθ, θ = cos^-1 x ∈ [0, π].
2. Integrand: tan^-1(tan(θ/2)) = θ/2.
3. Transform: I = ∫ θ2(-sinθ) dθ = -12∫ θ dθ.
4. Parts: ∫ θ dθ = -θ + sinθ (from Ex 7.6).
5. So I = -12[-θ + sinθ] + C = θ - sinθ2 + C.
6. Restore: θ = cos^-1 x, cosθ = x, sinθ = √1 - x²: I = xcos^-1 x - √1 - x²2 + C.
7. Tidier form: I = 12[xcos^-1 x - √1 - x²] + C.

12[xcos^-1 x - √1 - x²] + C

Concept used. Simplify the log: log(x² + 1) - 2log x = logx² + 1x² = log(1 + 1/x²). Then substitute t = 1 + 1/x², dt = -2/x³ dx.

1. Algebra: √x² + 1/x⁴ = 1x³√(x² + 1)/x² = 1x³√1 + 1/x².
2. So the integrand is 1x³√1 + 1/x² log(1 + 1/x²).
3. Let t = 1 + 1/x². Then dt = -2/x³ dx, so dx/x³ = -dt/2.
4. Transform: I = ∫ t log t · (-dt/2) = -12∫ t log t dt.
5. Parts on ∫ t^1/2log t dt: u = log t, dv = t^1/2 dt ⇒ du = dt/t, v = 2 t^3/23. ∫ t^1/2log t dt = 2 t^3/23log t - 23∫ t^1/2 dt = 2 t^3/23log t - 4 t^3/29.
6. Multiply by -1/2: I = -t^3/23log t + 2 t^3/29 + C = t^3/29[2 - 3log t] + C.
7. Restore t = 1 + 1/x² = (x² + 1)/x², so t^3/2 = (x² + 1)^3/2/x³: I = (x² + 1)^3/29 x³[2 - 3logx² + 1x²] + C.

-(x² + 1)^3/23 x³logx² + 1x² + 2(x² + 1)^3/29 x³ + C

Concept used. Half-angle: 1 - cos x = 2sin²(x/2) and 1 - sin x = 1 - 2sin(x/2)cos(x/2) = (sin(x/2) - cos(x/2))² · — no, simpler: split into two fractions.

1. Half-angle: 1 - sin x1 - cos x = 1 - 2sin(x/2)cos(x/2)2sin²(x/2) = 12sin²(x/2) - cos(x/2)sin(x/2) = 12cosec²(x/2) - cot(x/2).
2. Set f(x) = -cot(x/2). Then f'(x) = 12cosec²(x/2), so f(x) + f'(x) = 12cosec²(x/2) - cot(x/2), matching the bracket.
3. Apply template: ∫ e^x(f + f') dx = e^x f + C = -e^x cot(x/2).
4. Evaluate from π/2 to π: -e^π cot(π/2) - [-e^π/2cot(π/4)] = -e^π · 0 + e^π/2· 1 = e^π/2.

e^π/2

Concept used. Divide numerator and denominator by cos⁴ x, then substitute t = tan² x.

1. Divide top and bottom by cos⁴ x: sin xcos x / cos⁴ x1 + sin⁴ x / cos⁴ x = tan x sec² x1 + tan⁴ x.
2. Let t = tan² x, dt = 2tan x sec² x dx, so tan xsec² x dx = dt/2. Limits: x = 0 → t = 0; x = π/4 → t = 1.
3. Transform: I = ₀¹ dt/21 + t² = 12[tan^-1 t]₀¹ = 12·π4 = π8.

π/8

Concept used. Divide top and bottom by cos² x.

1. Rewrite the denominator using cos² + sin² = 1: cos² x + 4sin² x = (cos² x + sin² x) + 3sin² x = 1 + 3sin² x.
2. Use cos² x = 1 - sin² x in the numerator. The integrand becomes 1 - sin² x1 + 3sin² x.
3. Split: 1 - sin² x1 + 3sin² x = -13 + 4/31 + 3sin² x. (Verify: -13(1 + 3sin² x) + 43 = -13 - sin² x + 43 = 1 - sin² x. )
4. First piece: -13₀^π/2 dx = -π6.
5. Second piece: 43₀^π/2dx1 + 3sin² x. Substitute u = tan x, du = sec² x dx, sin² x = u²/(1 + u²), dx = du/(1 + u²). Then 1 + 3sin² x = (1 + u² + 3 u²)/(1 + u²) = (1 + 4 u²)/(1 + u²). So dx1 + 3sin² x = du/(1 + u²)(1 + 4 u²)/(1 + u²) = du1 + 4 u². Limits: x = 0 → u = 0; x = π/2 → u = ∞.
6. Integrate: 43₀^∞ du1 + 4 u² = 43· 12tan^-1(2 u)|₀^∞ = 23· π2 = π3.
7. Sum: I = -π6 + π3 = π6.

π/6

Concept used. Rewrite sin 2 x = 1 - (sin x - cos x)² and substitute u = sin x - cos x.

1. Identity: (sin x - cos x)² = sin² x - 2sin xcos x + cos² x = 1 - sin 2 x, so sin 2 x = 1 - u² where u = sin x - cos x.
2. Differentiate: du = (cos x + sin x) dx. So (sin x + cos x) dx = du.
3. Transform: I = ∫ du√1 - u².
4. Limits: x = π/6 → u = 1/2 - 3/2 = (1 - 3)/2; x = π/3 → u = 3/2 - 1/2 = (3 - 1)/2.
5. Antiderivative sin^-1 u: I = sin^-13 - 12 - sin^-11 - 32 = 2sin^-13 - 12. (Used sin^-1(-v) = -sin^-1 v.)

2sin^-13 - 12

Concept used. Rationalise.

1. Multiply by conjugate: 1√1 + x - x· √1 + x + x√1 + x + x = √1 + x + x(1 + x) - x = √1 + x + x.
2. Integrate piece by piece: I = ₀¹√1 + x dx + ₀¹x dx. Each is a power-rule integral with the substitution shift: I = [2(1 + x)^3/23]₀¹ + [2 x^3/23]₀¹.
3. Evaluate: 23(22 - 1) + 23(1) = 42 - 2 + 23 = 423.

423

Concept used. Substitute u = sin x - cos x (so du = (cos x + sin x) dx). Then sin 2 x = 1 - u², and the denominator becomes 9 + 16(1 - u²) = 25 - 16 u².

1. Transform: I = ∫ du25 - 16 u².
2. Limits: x = 0 → u = -1; x = π/4 → u = 0.
3. Factor: 25 - 16 u² = (5 - 4 u)(5 + 4 u). Use ∫ dua² - b² u² = 12 a blog|a + b ua - b u| with a = 5, b = 4: I = 140log|5 + 4 u5 - 4 u||_-1⁰.
4. At u = 0: log 1 = 0. At u = -1: log|(5 - 4)/(5 + 4)| = log(1/9) = -log 9.
5. Subtract: I = 140[0 - (-log 9)] = log 940 = 2log 340 = log 320.

log 320

Concept used. Substitute t = sin x, dt = cos x dx. Note sin 2 x = 2sin xcos x = 2 tcos x, so sin 2 x dx = 2 t dt.

1. Transform: I = ₀¹ 2 t tan^-1 t dt.
2. Parts: u = tan^-1 t, dv = 2 t dt ⇒ du = dt/(1 + t²), v = t². ∫ 2 t tan^-1 t dt = t² tan^-1 t - ∫ t²1 + t² dt.
3. ∫ t²1 + t² dt = ∫ (1 - 11 + t²) dt = t - tan^-1 t.
4. So antiderivative is t² tan^-1 t - t + tan^-1 t = (t² + 1)tan^-1 t - t.
5. Evaluate at t = 1: (2)· π/4 - 1 = π/2 - 1. At t = 0: 0.
6. I = π/2 - 1.

π2 - 1

Concept used. Three modulus integrands; split each at its breakpoint, integrate piecewise, then sum.

1. On [1, 4]: |x - 1| = x - 1 throughout (since x ≥ 1). |x - 2| = 2 - x for x ∈ [1, 2]; = x - 2 for x ∈ [2, 4]. |x - 3| = 3 - x for x ∈ [1, 3]; = x - 3 for x ∈ [3, 4].
2. ₁⁴ |x - 1| dx = ₁⁴ (x - 1) dx = [(x-1)²/2]₁⁴ = 9/2.
3. ₁⁴ |x - 2| dx = ₁²(2 - x) dx + ₂⁴(x - 2) dx = 1/2 + 2 = 5/2.
4. ₁⁴ |x - 3| dx = ₁³(3 - x) dx + ₃⁴(x - 3) dx = 2 + 1/2 = 5/2.
5. Sum: I = 9/2 + 5/2 + 5/2 = 19/2.

19/2

Concept used. Partial fractions with repeated linear factor.

1. Decompose: 1x²(x + 1) = Ax + Bx² + Cx + 1. Multiply: 1 = A x (x + 1) + B(x + 1) + C x². At x = 0: 1 = B ⇒ B = 1. At x = -1: 1 = C ⇒ C = 1. Coefficient of x²: 0 = A + C ⇒ A = -1.
2. Integrate: I = -log|x| - 1x + log|x + 1|.
3. Evaluate at x = 3: -log 3 - 1/3 + log 4.
   At x = 1: -log 1 - 1 + log 2 = -1 + log 2.
4. Difference: (-log 3 - 1/3 + log 4) - (-1 + log 2) = -log 3 + log 4 - log 2 + 1 - 1/3 = log(4/(3· 2)) + 2/3 = log(2/3) + 2/3.
5. This is the desired result.

23 + log23    QED.

Concept used. Integration by parts with u = x, dv = e^x dx.

1. u = x ⇒ du = dx. dv = e^x dx ⇒ v = e^x.
2. ∫ x e^x dx = x e^x - ∫ e^x dx = (x - 1)e^x.
3. Evaluate: F(1) - F(0) = (1 - 1)e - (-1)(1) = 0 + 1 = 1.

1    QED.

Concept used. Parity. x¹⁷ is odd; cos⁴ x is even; product is odd. Integral of an odd function over [-a, a] is 0.

1. Define f(x) = x¹⁷cos⁴ x. Check: f(-x) = (-x)¹⁷cos⁴(-x) = -x¹⁷cos⁴ x = -f(x). Odd.
2. By P₇(ii): _-1¹ f(x) dx = 0.

0    QED.

Concept used. Write sin³ x = sin x (1 - cos² x) and substitute t = cos x.

1. sin³ x = sin x - sin x cos² x.
2. ∫ sin x dx = -cos x. For the second piece, t = cos x, dt = -sin x dx: ∫ sin x cos² x dx = -∫ t² dt = -t³/3 = -cos³ x/3.
3. Antiderivative: -cos x + cos³ x/3.
4. Evaluate on [0, π/2]: at π/2: 0 + 0 = 0. At 0: -1 + 1/3 = -2/3. Difference: 0 - (-2/3) = 2/3.

2/3    QED.

Concept used. tan³ x = tan x(sec² x - 1) = tan x sec² x - tan x.

1. Split: 2tan³ x = 2tan xsec² x - 2tan x.
2. ∫ 2tan xsec² x dx: let u = tan x, du = sec² x dx; integral = u² = tan² x.
3. ∫ 2tan x dx = -2log|cos x| = 2log|sec x|.
4. Antiderivative: tan² x - 2log|sec x|.
5. Evaluate at π/4: 1 - 2log2 = 1 - log 2. At 0: 0 - 0 = 0.
6. Difference: 1 - log 2.

1 - log 2    QED.

Concept used. Parts with u = sin^-1 x, dv = dx.

1. du = dx/√1 - x², v = x.
2. ∫ sin^-1 x dx = xsin^-1 x - ∫ x dx√1 - x² = xsin^-1 x + √1 - x².
3. Evaluate at 1: 1· π/2 + 0 = π/2. At 0: 0 + 1 = 1.
4. Difference: π/2 - 1.

π/2 - 1    QED.

Concept used. Multiply top and bottom by e^x.

1. 1e^x + e^-x = e^xe^2x + 1.
2. Let t = e^x, dt = e^x dx: I = ∫ dt1 + t² = tan^-1 t + C = tan^-1(e^x) + C.
3. Matches (A).

Option (A): tan^-1(e^x) + C

Concept used. cos 2 x = cos² x - sin² x = (cos x - sin x)(cos x + sin x).

1. Cancel one factor of (sin x + cos x): cos 2 x(sin x + cos x)² = cos x - sin xsin x + cos x.
2. Note ddx(sin x + cos x) = cos x - sin x, which is the numerator.
3. Apply ∫ D'(x)D(x) dx = log|D(x)| + C: I = log|sin x + cos x| + C. Matches (B).

Option (B)

Concept used. Property P₃: _a^b f(x) dx = _a^b f(a + b - x) dx, applied to x f(x).

1. Let I = _a^b x f(x) dx. By P₃: I = _a^b (a + b - x) f(a + b - x) dx = _a^b (a + b - x) f(x) dx, using f(a + b - x) = f(x).
2. Add the two forms: 2 I = _a^b [x + (a + b - x)] f(x) dx = (a + b)_a^b f(x) dx.
3. Solve: I = a + b2_a^b f(x) dx. Matches (D).

Option (D)