Evaluate integral sinx/x dx as an infinite series.
The correct answer and explanation is :
To evaluate the integral ( \int \frac{\sin x}{x} \, dx ) as an infinite series, we can start by expressing ( \frac{\sin x}{x} ) as a Taylor series expansion and then integrate term by term.
Step 1: Taylor Series for ( \sin x )
The Taylor series expansion of ( \sin x ) around ( x = 0 ) is given by:
[
\sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdots
]
Step 2: Divide by ( x )
Now, divide the series for ( \sin x ) by ( x ) to obtain ( \frac{\sin x}{x} ):
[
\frac{\sin x}{x} = 1 – \frac{x^2}{3!} + \frac{x^4}{5!} – \frac{x^6}{7!} + \cdots
]
Step 3: Integrate Term by Term
Now, we can integrate ( \frac{\sin x}{x} ) term by term:
[
\int \frac{\sin x}{x} \, dx = \int \left( 1 – \frac{x^2}{3!} + \frac{x^4}{5!} – \frac{x^6}{7!} + \cdots \right) dx
]
Integrating each term individually:
[
\int 1 \, dx = x
]
[
\int \frac{x^2}{3!} \, dx = \frac{x^3}{3! \cdot 3} = \frac{x^3}{18}
]
[
\int \frac{x^4}{5!} \, dx = \frac{x^5}{5! \cdot 5} = \frac{x^5}{600}
]
[
\int \frac{x^6}{7!} \, dx = \frac{x^7}{7! \cdot 7} = \frac{x^7}{35280}
]
And so on.
Thus, the integral becomes:
[
\int \frac{\sin x}{x} \, dx = x – \frac{x^3}{18} + \frac{x^5}{600} – \frac{x^7}{35280} + \cdots + C
]
where ( C ) is the constant of integration.
Final Answer:
[
\int \frac{\sin x}{x} \, dx = x – \frac{x^3}{18} + \frac{x^5}{600} – \frac{x^7}{35280} + \cdots + C
]