Indefinite Integration
Indefinite Integration
nta_abhyas_2025
Grade None
Question:
Let $f(n, z) = \int \cos(nz) dx$, with $f(0, 0) = 0$. If the expression $\sum_{i=1}^{n} f(i, 1)$ simplifies to $\frac{\sin(a+b)}{\sin c}$, then the value of $\frac{a}{c}$ is (where $a > b$)
Step-by-Step Solution
Key Concept: Integrate to find the function, then use the sum-of-sines formula to evaluate the series.
$f(n,x) = \frac{\sin(nx)}{n} + C$. As $f(n,0) = C = 0$, we have $f(n,x) = \sin(nx)$. Thus, $\sum_{i=1}^{n} f(i,x) = \sin x + \sin 2x + \cdots + \sin(89x)$. Using the formula for sum of sines, this equals $\frac{\sin(\frac{n}{2}x)\sin(\frac{(n+1)}{2}x)}{\sin(\frac{x}{2})}$. Evaluating gives $\frac{\sin(\frac{1}{2})}{\sin(\frac{1}{4})}$.
Correct Answer: C