More practice questions

5. Q5. Find ∫x√(x + 2) dx (NCERT Exercise 7.2, Q7, page 240).

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   Put t = x + 2, so x = t − 2 and dx = dt. The integral is ∫(t − 2)t^1/2 dt = ∫(t^3/2 − 2t^1/2) dt = (2/5)t^5/2 − (4/3)t^3/2 + C, that is (2/5)(x + 2)^5/2 − (4/3)(x + 2)^3/2 + C.

6. Q6. Find ∫x/√(x + 4) dx, for x > 0 (NCERT Exercise 7.2, Q11, page 240).

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   Put t = x + 4: ∫(t − 4)t^−1/2 dt = (2/3)t^3/2 − 8t^1/2 + C. Taking out (2/3)√t leaves (2/3)√(x + 4)(x + 4 − 12), so the answer is (2/3)√(x + 4)(x − 8) + C.

7. Q7. Find ∫tan² (2x − 3) dx (NCERT Exercise 7.2, Q21, page 241).

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   tan² θ = sec² θ − 1, and the integral of sec² (2x − 3) is ½ tan (2x − 3). So the answer is ½ tan (2x − 3) − x + C.

8. Q8. Find ∫dx/(1 + cot x) (NCERT Exercise 7.2, Q32, page 241).

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   Keep the original domain: sin x ≠ 0 and sin x + cos x ≠ 0. There, 1/(1 + cot x) = sin x/(sin x + cos x). As in NCERT Example 6 (iii), write the top as half of (sin x + cos x) plus half of (sin x − cos x). The first half gives x/2. In the second, the top sin x − cos x is minus the derivative of sin x + cos x, which gives −½ log |sin x + cos x|. The answer is x/2 − ½ log |cos x + sin x| + C.

9. Q9. Find ∫sin⁴ x dx (NCERT Exercise 7.3, Q10, page 243).

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   sin⁴ x = [(1 − cos 2x)/2]² = ¼(1 − 2 cos 2x + cos² 2x), and cos² 2x = (1 + cos 4x)/2. So sin⁴ x = 3/8 − ½ cos 2x + (1/8) cos 4x, and the integral is 3x/8 − ¼ sin 2x + (1/32) sin 4x + C.

10. Q10. ∫(sin² x − cos² x)/(sin² x cos² x) dx is equal to (A) tan x + cot x + C; (B) tan x + cosec x + C; (C) −tan x + cot x + C; (D) tan x + sec x + C (NCERT Exercise 7.3, Q23, page 243).

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    Dividing each term of the top by the bottom, the integrand is sec² x − cosec² x, whose integral is tan x + cot x + C: (A).

11. Q11. ∫e^x(1 + x)/cos² (e^xx) dx equals (A) −cot (ex^x) + C; (B) tan (xe^x) + C; (C) tan (e^x) + C; (D) cot (e^x) + C (NCERT Exercise 7.3, Q24, page 243).

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    Put t = xe^x. By the product rule dt = e^x(1 + x) dx, so the integral is ∫sec² t dt = tan (xe^x) + C: (B). NCERT slip (page 243): option (A) is printed −cot (ex^x), probably for −cot (xe^x). It is wrong either way: the derivative of −cot (xe^x) is cosec² (xe^x) × e^x(1 + x). In an exam, test each option by differentiating it.

12. Q12. Find ∫sin^⁻¹ (cos x) dx (NCERT Exercise 7.3, Q21, page 243). On which interval does the answer hold?

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    cos x = sin (π/2 − x), and sin^⁻¹ (sin θ) = θ only for θ in [−π/2, π/2], that is for x in [0, π]. There sin^⁻¹ (cos x) = π/2 − x, and the integral is πx/2 − x²/2 + C, NCERT’s answer. The question gives no interval; outside [0, π] the integrand is a different function (for x in [−π, 0] it is π/2 + x), so this formula does not hold there. Beyond the NCERT text (JEE): check the principal branch before simplifying.