Class 12 Maths Chapter 7: Integrals worksheet with answer key

Answer: (b) 14 sin 4x + e^−x + C

ddx(14 sin 4x) = cos 4x and ddx(e^−x) = − e^−x, so the integral is 14 sin 4x + e^−x + C.

Answer: (b) 15 tan^-1 x5 + C

∫ dxx² + a² = 1a tan^-1 xa + C with a = 5. Option (d) is the form for x² − 25.

Answer: (c) log |cosec x − cot x| + C

Multiplying and dividing by (cosec x − cot x) and putting cosec x − cot x = t gives ∫ cosec x dx = log |cosec x − cot x| + C. Option (d) is ∫ cot x dx; (a) is the derivative of cosec x.

Answer: (c) 14

F(x) = 2x³ − 2x² + x + C. F(1) = 1 + C = 5 gives C = 4, so F(2) = 16 − 8 + 2 + 4 = 14.

Answer: (b) sec x + x² + C

d/dx (sec x) = sec x tan x and d/dx (x²) = 2x, so ∫ (sec x tan x + 2x) dx = sec x + x² + C.

y = ∫ (3x² − 2x⁻²) dx = x³ + 2x + C. At (1, 4): 4 = 1 + 2 + C, so C = 1 and y = x³ + 2x + 1.

• y = x³ + 2x + C (1)
• uses (1, 4) to get C = 1 and states the curve (1)

Put e^x + x = t; then (e^x + 1) dx = dt and the integral is ∫ dtt = log |e^x + x| + C.

• substitution e^x + x = t with dt = (e^x + 1) dx (1)
• log |e^x + x| + C (1)

Put t = 1 + 3x², dt = 6x dx; x = 0 gives t = 1 and x = 1 gives t = 4. The integral is 16 ∫₁⁴ t^−1/2 dt = 16 [2√t]₁⁴ = 16(4 − 2) = 13.

• substitution with the limits changed to 1 and 4 (1)
• value 13 (1)

sin² x cos² x = 14 sin² 2x = 18(1 − cos 4x). Hence ∫ sin² x cos² x dx = x8 − sin 4x32 + C.

• writes the integrand as 14 sin² 2x (1)
• uses sin² θ = 1 − cos 2θ2 to get 18(1 − cos 4x) (1)
• x8 − sin 4x32 + C (1)

Let I be the integral. By P₃ with a + b = 3, replacing x by 3 − x: I = ∫₁² e^{3 − x}e^{3 − x} + e^x dx. Adding the two forms, 2I = ∫₁² 1 dx = 1, so I = 12.

• applies P₃ to get the second form of I (1)
• adds to get 2I = ∫₁² dx (1)
• concludes I = 12 (1)

8 + 2x − x² = 9 − (x − 1)². Put x − 1 = t; t runs from 0 to 32. ∫₀^3/2 √(9 − t²) dt = [t2 √(9 − t²) + 92 sin^-1 t3]₀^3/2 = 34 · 3√32 + 92 · π6 = 9√38 + 3π4.

• completes the square to 9 − (x − 1)² (1)
• uses the formula for ∫ √(a² − t²) dt with the new limits (1)
• value 9√38 + 3π4 (1)