Maclaurin Series For Sin^2(x) & Integral Evaluation

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a problem that combines the elegance of Maclaurin series expansions with the practical application of definite integrals. We're going to unravel the first three non-zero terms in the Maclaurin series expansion of $f(x) = \sin^2 x$ and then use this powerful tool to evaluate a tricky integral: $\int_0^1 \frac{\sin^2 x}{\sqrt{x}} dx$, accurate to three decimal places. This isn't just about crunching numbers; it's about understanding how we can approximate complex functions and use those approximations to solve problems that might otherwise be intractable. So, buckle up, grab your thinking caps, and let's get started on this mathematical adventure!

Unpacking the Maclaurin Series for $f(x) = \sin^2 x$

The Maclaurin series is a special case of the Taylor series, representing a function as an infinite sum of terms calculated from the values of its derivatives at a single point, typically $x=0$. For a function $f(x)$, the Maclaurin series is given by:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots + \frac{f^{(n)}(0)}{n!}x^n + \dots$

Our goal here is to find the first three non-zero terms for $f(x) = \sin^2 x$. We could go the long route of calculating derivatives of $\sin^2 x$ repeatedly, but that can get messy. A smarter approach is to use known Maclaurin series. We all know the Maclaurin series for $\sin x$ is:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

Now, we need to square this series to get the series for $\sin^2 x$.

$\sin^2 x = \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right)^2$

Let's expand this. We'll focus on the first few terms to ensure we get the first three non-zero terms for $\sin^2 x$.

$\sin^2 x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \right) \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \right)$

When we multiply these out, we'll take terms that give us powers of $x$.

• The first term will be $x \times x = x^2$. This is definitely non-zero.
• The next terms will involve $x$ multiplied by terms from the second bracket, and terms from the first bracket multiplied by $x$. Let's look for terms that result in $x^4$ or $x^5$ to find subsequent non-zero terms.

When we multiply the first two terms of each series:

$(x - \frac{x^3}{6}) (x - \frac{x^3}{6}) = x^2 - \frac{x^4}{6} - \frac{x^4}{6} + \frac{x^6}{36} = x^2 - \frac{2x^4}{6} + \frac{x^6}{36} = x^2 - \frac{x^4}{3} + \frac{x^6}{36}$

Now let's consider the $x^5$ term from the original $\sin x$ series. Squaring the first term $(x)$ gives us $x^2$. Squaring the second term $(-\frac{x^3}{6})$ gives us $\frac{x^6}{36}$. Multiplying the first term $(x)$ by the third term $(\frac{x^5}{120})$ and vice versa gives us $2 \times x \times \frac{x^5}{120} = \frac{2x^6}{120} = \frac{x^6}{60}$.

This direct squaring can get complicated quickly. There's a much cleaner way using trigonometric identities! Recall the double angle identity: $\cos(2x) = 1 - 2\sin^2 x$. Rearranging this, we get:

$\sin^2 x = \frac{1 - \cos(2x)}{2}$

This is a game-changer, guys! Now we just need the Maclaurin series for $\cos(2x)$. We know the Maclaurin series for $\cos u$ is:

$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \dots$

Let $u = 2x$. Substituting $2x$ for $u$ gives us:

$\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \dots$

$\cos(2x) = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \frac{64x^6}{720} + \dots$

$\cos(2x) = 1 - 2x^2 + \frac{2}{3}x^4 - \frac{4}{45}x^6 + \dots$

Now, substitute this back into our expression for $\sin^2 x$:

$\sin^2 x = \frac{1 - \left( 1 - 2x^2 + \frac{2}{3}x^4 - \frac{4}{45}x^6 + \dots \right)}{2}$

$\sin^2 x = \frac{1 - 1 + 2x^2 - \frac{2}{3}x^4 + \frac{4}{45}x^6 - \dots}{2}$

$\sin^2 x = \frac{2x^2 - \frac{2}{3}x^4 + \frac{4}{45}x^6 - \dots}{2}$

$\sin^2 x = x^2 - \frac{1}{3}x^4 + \frac{2}{45}x^6 - \dots$

So, the first three non-zero terms in the Maclaurin series expansion of $f(x) = \sin^2 x$ are $x^2$, $-\frac{1}{3}x^4$, and $\frac{2}{45}x^6$. This is a much cleaner and more reliable way to get there, proving that knowing your trig identities can save you a ton of calculus headaches!

Evaluating the Integral Using the Maclaurin Series

Alright guys, now for the second part of the challenge: evaluating the integral $\int_0^1 \frac{\sin^2 x}{\sqrt{x}} dx$ correct to three decimal places. This is where our Maclaurin series for $\sin^2 x$ comes into play. Since the integral is from 0 to 1, and our Maclaurin series expansion is valid around $x=0$, we can substitute the series expansion into the integral.

We found that $\sin^2 x = x^2 - \frac{1}{3}x^4 + \frac{2}{45}x^6 - \dots$.

So, the integrand becomes:

$\frac{\sin^2 x}{\sqrt{x}} = \frac{x^2 - \frac{1}{3}x^4 + \frac{2}{45}x^6 - \dots}{x^{1/2}}$

We can divide each term by $x^{1/2}$:

$\frac{\sin^2 x}{\sqrt{x}} = x^{2 - 1/2} - \frac{1}{3}x^{4 - 1/2} + \frac{2}{45}x^{6 - 1/2} - \dots$

$\frac{\sin^2 x}{\sqrt{x}} = x^{3/2} - \frac{1}{3}x^{7/2} + \frac{2}{45}x^{11/2} - \dots$

Now, we need to integrate this series term by term from 0 to 1:

$\int_0^1 \frac{\sin^2 x}{\sqrt{x}} dx = \int_0^1 \left( x^{3/2} - \frac{1}{3}x^{7/2} + \frac{2}{45}x^{11/2} - \dots \right) dx$

We can integrate each term using the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$.

$\int_0^1 x^{3/2} dx = \left[ \frac{x^{3/2 + 1}}{3/2 + 1} \right]_0^1 = \left[ \frac{x^{5/2}}{5/2} \right]_0^1 = \left[ \frac{2}{5}x^{5/2} \right]_0^1 = \frac{2}{5}(1)^{5/2} - \frac{2}{5}(0)^{5/2} = \frac{2}{5}$

$\int_0^1 -\frac{1}{3}x^{7/2} dx = -\frac{1}{3} \left[ \frac{x^{7/2 + 1}}{7/2 + 1} \right]_0^1 = -\frac{1}{3} \left[ \frac{x^{9/2}}{9/2} \right]_0^1 = -\frac{1}{3} \left[ \frac{2}{9}x^{9/2} \right]_0^1 = -\frac{1}{3} \left( \frac{2}{9}(1)^{9/2} - \frac{2}{9}(0)^{9/2} \right) = -\frac{1}{3} \times \frac{2}{9} = -\frac{2}{27}$

$\int_0^1 \frac{2}{45}x^{11/2} dx = \frac{2}{45} \left[ \frac{x^{11/2 + 1}}{11/2 + 1} \right]_0^1 = \frac{2}{45} \left[ \frac{x^{13/2}}{13/2} \right]_0^1 = \frac{2}{45} \left[ \frac{2}{13}x^{13/2} \right]_0^1 = \frac{2}{45} \left( \frac{2}{13}(1)^{13/2} - \frac{2}{13}(0)^{13/2} \right) = \frac{2}{45} \times \frac{2}{13} = \frac{4}{585}$

So, the integral is approximately:

$\int_0^1 \frac{\sin^2 x}{\sqrt{x}} dx \approx \frac{2}{5} - \frac{2}{27} + \frac{4}{585} - \dots$

Let's calculate these values:

• $\frac{2}{5} = 0.4$
• $\frac{2}{27} \approx 0.074074...$
• $\frac{4}{585} \approx 0.006837...$

Now, let's sum the first three terms:

$0.4 - 0.074074 + 0.006837 approx 0.332763$

To be confident about the accuracy to three decimal places, we should consider if the next term would significantly alter the result. The next term in the Maclaurin series for $\sin^2 x$ involves $x^8$. When divided by $\sqrt{x}$, this would be $x^{8 - 1/2} = x^{15/2}$. Integrating this from 0 to 1 gives $\frac{2}{17}$. The coefficient of $x^8$ in $\sin^2 x$ is derived from $(\frac{-4}{45}x^6)(x^2)$ and $(x^2)(\frac{-4}{45}x^6)$ and others. This can get complex.

Alternatively, let's use the established identity $\sin^2 x = 1 - \cos(2x) / 2$. The Maclaurin series for $\cos(2x)$ is $1 - 2x^2 + \frac{2}{3}x^4 - \frac{4}{45}x^6 + \frac{2}{315}x^8 - \dots$. So $\sin^2 x = x^2 - \frac{1}{3}x^4 + \frac{2}{45}x^6 - \frac{1}{315}x^8 + \dots$.

The integral of the next term $(-\frac{1}{315}x^8)$ divided by $\sqrt{x}$ is $\int_0^1 -\frac{1}{315}x^{15/2} dx = -\frac{1}{315} \left[\frac{x^{17/2}}{17/2}\right]_0^1 = -\frac{1}{315} \times \frac{2}{17} = -\frac{2}{5355}$.

$\frac{2}{5355} \approx 0.000373...$

Our current sum is $0.332763$. The next term would subtract approximately $0.000373$.

$0.332763 - 0.000373 \approx 0.33239$

This means our sum $0.332763$ is slightly too high. If we round $0.332763$ to three decimal places, we get $0.333$. However, considering the effect of the next term, $0.33239$ rounded to three decimal places is $0.332$.

Let's re-evaluate the sum with higher precision:

$\frac{2}{5} = 0.4$

$\frac{2}{27} \approx 0.07407407$

$\frac{4}{585} \approx 0.00683760$

Sum of first three terms: $0.4 - 0.07407407 + 0.00683760 \approx 0.33276353$

Term 4: $-\frac{2}{5355} \approx -0.00037340$

Sum of first four terms: $0.33276353 - 0.00037340 \approx 0.33239013$

Rounding $0.33239013$ to three decimal places gives us 0.332. It's crucial to check the next term to ensure accuracy, especially when dealing with approximations!

Conclusion

And there you have it, math enthusiasts! We've successfully navigated the realms of Maclaurin series expansions and definite integrals. By cleverly using trigonometric identities, we found the first three non-zero terms of $\sin^2 x$'s Maclaurin series to be $x^2$, $-\frac{1}{3}x^4$, and $\frac{2}{45}x^6$. Then, we employed this series to approximate the integral $\int_0^1 \frac{\sin^2 x}{\sqrt{x}} dx$. The journey involved term-by-term integration and careful summation, reminding us of the importance of checking subsequent terms for precision. We arrived at a value of approximately 0.332 correct to three decimal places. This problem beautifully illustrates the power of series expansions in tackling complex analytical problems. Keep practicing, keep exploring, and I'll catch you in the next one!