Definite Integration
Reduction formulae
Grade Class 12
Question:
If I_n = ∫(sin x)^n dx n ∈ N, then 5I_4 - 6I_6 is equal to -
(A) sin x.(cos x)^5 + C
(B) cos x.(sin x)^5 + C
(C) \frac{\sin 2x}{8}[\cos^2 2x + 1 - 2\cos 2x] + C
(D) \frac{\sin 2x}{8}[\cos^2 2x + 1 + 2\cos 2x] + C
Step-by-Step Solution
Key Concept: Use the reduction formula for I_n = \int(sin x)^n dx, which is I_n = -(1/n)(sin x)^(n-1)cos x + ((n-1)/n)I_(n-2). Apply this to find the relation between I_6 and I_4.
The reduction formula for I_n = \int(sin x)^n dx is I_n = -(1/n)(sin x)^(n-1)cos x + ((n-1)/n)I_(n-2). For n=6, I_6 = -(1/6)(sin x)^5 cos x + (5/6)I_4. Rearranging gives 6I_6 = -(sin x)^5 cos x + 5I_4, which implies 5I_4 - 6I_6 = (sin x)^5 cos x + C. Further manipulation of trigonometric identities shows that options (B) and (C) are equivalent.
Correct Answer: 2, 3