Definite Integration
Integration of trigonometric functions
Grade Class 12
Question:
Suppose J = ∫ \frac{\sin^2 x + \sin x}{1 + \sin x + \cos x} dx and K = ∫ \frac{\cos^2 x + \cos x}{1 + \sin x + \cos x} dx. If C is an arbitrary constant of integration then which of the following is/are correct?
(A) J = \frac{1}{2} (x - \sin x + \cos x) + C
(B) J = K - (\sin x + \cos x) + C
(C) J = x - K + C
(D) K = \frac{1}{2} (x - \sin x + \cos x) + C
Step-by-Step Solution
Key Concept: Calculate J+K and J-K. J+K = \int (sin^2 x + cos^2 x + sin x + cos x) / (1 + sin x + cos x) dx = \int (1 + sin x + cos x) / (1 + sin x + cos x) dx = \int 1 dx = x + C. J-K = \int (sin^2 x - cos^2 x + sin x - cos x) / (1 + sin x + cos x) dx = \int ((sin x - cos x)(sin x + cos x) + (sin x - cos x)) / (1 + sin x + cos x) dx = \int (sin x - cos x)(sin x + cos x + 1) / (1 + sin x + cos x) dx = \int (sin x - cos x) dx = -cos x - sin x + C.
J + K = \int \frac{\sin^2 x + \cos^2 x + \sin x + \cos x}{1 + \sin x + \cos x} dx = \int \frac{1 + \sin x + \cos x}{1 + \sin x + \cos x} dx = x + C_1. J - K = \int \frac{\sin^2 x - \cos^2 x + \sin x - \cos x}{1 + \sin x + \cos x} dx = \int \frac{(\sin x - \cos x)(\sin x + \cos x + 1)}{1 + \sin x + \cos x} dx = \int (\sin x - \cos x) dx = -\cos x - \sin x + C_2. Adding the two equations: 2J = x - \sin x - \cos x + C, so J = \frac{1}{2}(x - \sin x - \cos x) + C. Subtracting the two equations: 2K = x + \sin x + \cos x + C, so K = \frac{1}{2}(x + \sin x + \cos x) + C. Checking options: (A) J = \frac{1}{2}(x - \sin x - \cos x) + C (Incorrect sign for cos x). (B) J - K = -\sin x - \cos x + C (Correct). (C) J + K = x + C (Correct). (D) K = \frac{1}{2}(x + \sin x + \cos x) + C (Incorrect sign for sin x and cos x).
Correct Answer: 1, 3