More practice questions

5. Q5. Find ∫tan^⁻¹ x dx (NCERT Exercise 7.6, Q13, page 263).

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   Take 1 as the second function: x tan^⁻¹ x − ∫x/(1 + x²) dx = x tan^⁻¹ x − ½ log (1 + x²) + C.

6. Q6. Find ∫e^2x sin x dx (NCERT Exercise 7.6, Q21, page 264).

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   Let I be the integral. With e^2x first twice: I = −e^2x cos x + 2∫e^2x cos x dx = −e^2x cos x + 2e^2x sin x − 4I. So 5I = e^2x(2 sin x − cos x), and I = (e^2x/5)(2 sin x − cos x) + C. Beyond the NCERT text, the same two steps give the integral of e^ax sin bx as e^ax(a sin bx − b cos bx)/(a² + b²) + C, valid for a² + b² > 0; if a = b = 0, the integral is C; a = 2, b = 1 gives the answer above.

7. Q7. Find ∫(sin^⁻¹ x)² dx (NCERT Exercise 7.6, Q10, page 263).

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   With 1 as the second function: x(sin^⁻¹ x)² − 2∫x sin^⁻¹ x/√(1 − x²) dx. The last integral is NCERT Example 20, x − √(1 − x²) sin^⁻¹ x. So the answer is x(sin^⁻¹ x)² + 2√(1 − x²) sin^⁻¹ x − 2x + C.

8. Q8. Find ∫e^x(1/x − 1/x²) dx (NCERT Exercise 7.6, Q19, page 264).

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   With f(x) = 1/x, f′(x) = −1/x², so the integral is e^x/x + C.

9. Q9. Find ∫(x − 3)e^x/(x − 1)³ dx (NCERT Exercise 7.6, Q20, page 264).

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   Write x − 3 = (x − 1) − 2: the fraction is 1/(x − 1)² − 2/(x − 1)³, which is f + f′ with f(x) = 1/(x − 1)². The integral is e^x/(x − 1)² + C.

10. Q10. ∫e^x sec x (1 + tan x) dx equals (A) e^x cos x + C; (B) e^x sec x + C; (C) e^x sin x + C; (D) e^x tan x + C (NCERT Exercise 7.6, Q24, page 264).

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    sec x(1 + tan x) = sec x + sec x tan x, which is f + f′ with f(x) = sec x. So the integral is e^x sec x + C: (B).

11. Q11. Find ∫√(x² + 4x + 6) dx (NCERT Exercise 7.7, Q3, page 266).

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    x² + 4x + 6 = (x + 2)² + (√2)². By the second square-root formula with x + 2 for x and a² = 2: ((x + 2)/2)√(x² + 4x + 6) + log |x + 2 + √(x² + 4x + 6)| + C.

12. Q12. Find ∫√(1 − 4x − x²) dx (NCERT Exercise 7.7, Q5, page 266).

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    1 − 4x − x² = 5 − (x + 2)². By the third square-root formula with a² = 5: ((x + 2)/2)√(1 − 4x − x²) + (5/2) sin^⁻¹ ((x + 2)/√5) + C.

13. Q13. ∫√(1 + x²) dx is equal to (A) (x/2)√(1 + x²) + ½ log |x + √(1 + x²)| + C; (B) (2/3)(1 + x²)^3/2 + C; (C) (2/3)x(1 + x²)^3/2 + C; (D) (x²/2)√(1 + x²) + ½x² log |x + √(1 + x²)| + C (NCERT Exercise 7.7, Q10, page 266).

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    The second square-root formula with a = 1 gives (A). (B) is the trap: it treats √(1 + x²) like a power of x, but its derivative is 2x√(1 + x²), not √(1 + x²).