Mastering Integrals: Your Guide To A Tricky Calculus Problem

Hey calculus enthusiasts! Ever stared at an integral and just felt… stuck? Yeah, me too. Today, we're diving deep into a particularly juicy one: evaluating the integral of $\frac{5 - 2x}{\sqrt{16x^2 + 25}} \, dx$. This might look a bit intimidating with that square root and the binomial in the numerator, but trust me, guys, it's totally doable once you break it down. We're going to walk through this step-by-step, making sure you understand the why behind each move. So, grab your favorite study snack, settle in, and let's conquer this integral together!

Deconstructing the Integral: The First Big Step

The first thing you probably noticed, just like our user did, is that the numerator has two terms: a constant (5) and a term with x (-2x). This is a huge hint! Whenever you see a sum or difference in the numerator of an integral, especially when it involves a square root in the denominator, the smartest play is often to split the integral into separate, more manageable pieces. Our user correctly identified this, breaking down the integral $I = \int \frac{5 - 2x}{\sqrt{16x^2 + 25}} \, dx$ into two parts:

$$I = \int \frac{5}{\sqrt{16x^2 + 25}} \, dx - \int \frac{2x}{\sqrt{16x^2 + 25}} \, dx$$

This is a brilliant starting point because each of these new integrals is likely to be solvable using standard integration techniques. We've transformed one potentially complex problem into two problems that look much more familiar. This strategy of decomposition is super powerful in calculus. It's all about simplifying the beast into smaller, more bite-sized creatures. So, high fives all around for getting to this stage! Now, let's tackle each of these beasts individually. We'll call the first one $I_1$ and the second one $I_2$. Get ready, because we're about to get our hands dirty with some serious integration.

Tackling the First Part: $I_1 = \int \frac{5}{\sqrt{16x^2 + 25}} \, dx$

Alright team, let's focus on the first integral: $I_1 = \int \frac{5}{\sqrt{16x^2 + 25}} \, dx$. The constant '5' can just be pulled right out, so we're looking at $5 \int \frac{1}{\sqrt{16x^2 + 25}} \, dx$. Now, this form, $ \frac{1}{\sqrt{ax^2 + b^2}} $, should be ringing some alarm bells. This is a classic case where trigonometric substitution is your best friend. We need to make the expression under the square root look like a perfect square of a trigonometric function. Specifically, we're aiming for something like $\tan^2(\theta) + 1 = \sec^2(\theta)$.

To get there, we first need to factor out the 16 from inside the square root: $\sqrt{16(x^2 + \frac{25}{16})}$. This becomes $4\sqrt{x^2 + (\frac{5}{4})^2}$. So, our integral $I_1$ is now $5 \int \frac{1}{4\sqrt{x^2 + (\frac{5}{4})^2}} \, dx = \frac{5}{4} \int \frac{1}{\sqrt{x^2 + (\frac{5}{4})^2}} \, dx$.

Now, we make the substitution. We want $x^2 + (\frac{5}{4})^2$ to become $(\frac{5}{4})^2 \tan^2(\theta) + (\frac{5}{4})^2$. This means we should let $x = \frac{5}{4} \tan(\theta)$. If we do that, then $dx = \frac{5}{4} \sec^2(\theta) \, d\theta$.

Let's substitute these into the integral:

$$\frac{5}{4} \int \frac{1}{\sqrt{(\frac{5}{4} \tan(\theta))^2 + (\frac{5}{4})^2}} \cdot \frac{5}{4} \sec^2(\theta) \, d\theta$$

Simplify inside the square root: $ \sqrt{(\frac{5}{4})^2 \tan^2(\theta) + (\frac{5}{4})^2} = \sqrt{(\frac{5}{4})^2 (\tan^2(\theta) + 1)} = \sqrt{(\frac{5}{4})^2 \sec^2(\theta)} = \frac{5}{4} \sec(\theta) $.

Plugging this back in:

$$\frac{5}{4} \int \frac{1}{\frac{5}{4} \sec(\theta)} \cdot \frac{5}{4} \sec^2(\theta) \, d\theta$$

The $ \frac{5}{4} $ terms cancel out nicely, and we're left with:

$$\frac{5}{4} \int \sec(\theta) \, d\theta$$

And the integral of $\sec(\theta)$ is a standard result: $ \ln|\sec(\theta) + \tan(\theta)| $. So, we have:

$$I_1 = \frac{5}{4} \ln|\sec(\theta) + \tan(\theta)| + C_1$$

But we're not done! We need to substitute back in terms of $x$. Remember our substitution: $x = \frac{5}{4} \tan(\theta)$. This means $\tan(\theta) = \frac{4x}{5}$. We can visualize this with a right triangle. If $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{4x}{5}$, then the opposite side is $4x$ and the adjacent side is $5$. The hypotenuse, by the Pythagorean theorem, is $\sqrt{(4x)^2 + 5^2} = \sqrt{16x^2 + 25}$.

Now we can find $\sec(\theta)$. Since $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\text{hypotenuse}}{\text{adjacent}}$, we get $\sec(\theta) = \frac{\sqrt{16x^2 + 25}}{5}$.

Substituting these back into our expression for $I_1$:

$$I_1 = \frac{5}{4} \ln\left|\frac{\sqrt{16x^2 + 25}}{5} + \frac{4x}{5}\right| + C_1$$

We can simplify the expression inside the logarithm: $ \frac{1}{5}(\sqrt{16x^2 + 25} + 4x) $. So:

$$I_1 = \frac{5}{4} \ln\left|\frac{\sqrt{16x^2 + 25} + 4x}{5}\right| + C_1$$

Using logarithm properties ($ \ln(a/b) = \ln(a) - \ln(b) $), we can pull out the $ \ln(5) $ term, which gets absorbed into the constant $C_1$. So, a cleaner form is:

$$I_1 = \frac{5}{4} \ln|\sqrt{16x^2 + 25} + 4x| + C_1$$

Phew! That was a workout, but we nailed the first part. Remember, trigonometric substitution is key for those $ \sqrt{ax^2 + b^2} $ forms.

Tackling the Second Part: $I_2 = \int \frac{2x}{\sqrt{16x^2 + 25}} \, dx$

Now, let's move on to the second integral, $I_2 = \int \frac{2x}{\sqrt{16x^2 + 25}} \, dx$. This one looks way simpler, right? And it is! This is a prime candidate for u-substitution. Look at the expression inside the square root: $16x^2 + 25$. If we differentiate that, we get $32x$. And guess what? We have an $x$ in the numerator! This is a perfect match.

Let $u = 16x^2 + 25$. Then, differentiate both sides with respect to $x$: $ \frac{du}{dx} = 32x $. Rearranging this, we get $ du = 32x , dx $.

Now, let's look back at our integral $I_2$. We have $2x \, dx$ in the numerator. We can rewrite $du = 32x \, dx$ as $ \frac{1}{16} du = 2x , dx $. Perfect!

Let's substitute $u$ and $ \frac{1}{16} du $ into the integral:

$$I_2 = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{16} du$$

We can pull the constant $ \frac{1}{16} $ out:

$$I_2 = \frac{1}{16} \int \frac{1}{\sqrt{u}} \, du$$

Rewrite $ \frac{1}{\sqrt{u}} $ as $ u^{-1/2} $:

$$I_2 = \frac{1}{16} \int u^{-1/2} \, du$$

Now, we use the power rule for integration: $ \int u^n , du = \frac{u^{n+1}}{n+1} + C $. In our case, $n = -1/2$, so $n+1 = 1/2$.

$$I_2 = \frac{1}{16} \cdot \frac{u^{1/2}}{1/2} + C_2$$

Simplifying this gives:

$$I_2 = \frac{1}{16} \cdot 2 u^{1/2} + C_2 = \frac{1}{8} u^{1/2} + C_2$$

Remember $u = 16x^2 + 25$. Substitute back:

$$I_2 = \frac{1}{8} (16x^2 + 25)^{1/2} + C_2 = \frac{1}{8} \sqrt{16x^2 + 25} + C_2$$

See? That wasn't so bad! U-substitution is incredibly handy when you have a function and its derivative (or a constant multiple of it) lurking in the integrand. It really simplifies things down to basic power rules.

Putting It All Together: The Final Solution

Now for the grand finale, guys! We have successfully evaluated both parts of our original integral. Recall that $I = I_1 - I_2$. So, we just need to subtract our result for $I_2$ from our result for $I_1$.

$$I = \left( \frac{5}{4} \ln|\sqrt{16x^2 + 25} + 4x| + C_1 \right) - \left( \frac{1}{8} \sqrt{16x^2 + 25} + C_2 \right)$$

Combining the constants $C_1$ and $C_2$ into a single constant $C$, we get our final answer:

$$I = \frac{5}{4} \ln|\sqrt{16x^2 + 25} + 4x| - \frac{1}{8} \sqrt{16x^2 + 25} + C$$

And there you have it! We've conquered that challenging integral by breaking it down into two simpler problems, using trigonometric substitution for one and u-substitution for the other. This problem really showcases the power of having a toolkit of different integration techniques and knowing when to apply them. Remember, the key is often to simplify, decompose, and substitute. Keep practicing, and you'll be a calculus wizard in no time!