Definite Integration
Indefinite Integration
Grade Class 12
Question:
The integral ∫(sin(101x)·sin^99 x)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 substitute into the integral and use integration by parts or recognize the derivative of the product.
Let I = \int sin(101x) sin^99 x 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^99 x dx = \int sin(100x) cos x sin^99 x dx + \int cos(100x) sin^100 x dx. Integrating the first part by parts, taking sin^100 x as the function to integrate and sin(100x) as the function to differentiate, we get \int sin(100x) sin^99 x cos x dx = sin(100x) * (sin^100 x / 100) - \int 100 cos(100x) * (sin^100 x / 100) dx = (sin(100x) sin^100 x) / 100 - \int cos(100x) sin^100 x dx. Substituting this back into the expression for I, the integral \int cos(100x) sin^100 x dx cancels out, leaving I = (sin(100x) sin^100 x) / 100 + C.
Correct Answer: A