Definite Integration
Reduction Formula
Grade Class 12
Question:
If I<sub>n</sub> = ∫(sin x)<sup>n</sup> dx, n ∈ N, then 5I<sub>4</sub> - 6I<sub>6</sub> is equal to -
(A) sin x.(cos x)<sup>5</sup> + C
(B) cos x.(sin x)<sup>5</sup> + C
(C) <sup>sin 2x</sup>/<sub>8</sub> [cos<sup>2</sup> 2x + 1 - 2cos 2x] + C
(D) <sup>sin 2x</sup>/<sub>8</sub> [cos<sup>2</sup> 2x + 1 + 2cos 2x] + C
Step-by-Step Solution
Key Concept: Use the reduction formula for integral of (sin x)^n dx: I_n = -(1/n)(sin x)^(n-1) cos x + ((n-1)/n) I_(n-2).
We know the reduction formula for I<sub>n</sub> = ∫(sin x)<sup>n</sup> dx is I<sub>n</sub> = -(1/n)(sin x)<sup>n-1</sup> cos x + ((n-1)/n) I<sub>n-2</sub>. Thus, nI<sub>n</sub> = -(sin x)<sup>n-1</sup> cos x + (n-1)I<sub>n-2</sub>. Rearranging gives (n-1)I<sub>n-2</sub> - nI<sub>n</sub> = (sin x)<sup>n-1</sup> cos x. For n=6, we have 5I<sub>4</sub> - 6I<sub>6</sub> = (sin x)<sup>5</sup> cos x + C. This matches option (B). Option (C) is an equivalent trigonometric form.
Correct Answer: B,C