Definite Integration
Integration of trigonometric functions
Grade Class 12
Question:
13. ∫(sin(101x)·sin⁹⁹x)dx equals
(A) \frac{\sin(100x)(\sin x)^{100}}{100} + C
(B) \frac{\cos(100x)(\sin x)^{100}}{100} + C
(C) \frac{\cos(100x)(\cos x)^{100}}{100} + C
(D) \frac{\sin(100x)(\sin x)^{101}}{101} + C
Step-by-Step Solution
Key Concept: Use the trigonometric identity sin(A+B) = sin A cos B + cos A sin B to expand sin(101x) as sin(100x+x) = sin(100x)cos x + cos(100x)sin x. Then use the substitution method or integration by parts.
Let I = \intsin(101x)sin^9^9x dx. Using sin(101x) = sin(100x+x) = sin(100x)cos x + cos(100x)sin x, we get I = \int(sin(100x)cos x + cos(100x)sin x)sin^9^9x dx = \intsin(100x)cos x sin^9^9x dx + \intcos(100x)sin^1^0^0x dx. This can be solved by recognizing the derivative of sin(100x)sin^1^0^0x.
Correct Answer: 1