Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{2}{25z^2+25}dz$$

Short answer

Question: Find the indefinite integral of the function $f(z)=\frac{2}{25z^2+25}$. Answer: The indefinite integral of the function $f(z)=\frac{2}{25z^2+25}$ is: $\int f(z)dz=\frac{2}{25}\arctan (z)+C$, where $C$ is the constant of integration.

Step-by-step solution

Recognize the form of the function

The given function is a rational function, meaning it is a ratio of two polynomials. Notice that the denominator is a perfect square plus a constant. With that in mind, let's consider the substitution method to simplify the integration process.

Apply substitution

Let's find a substitution that simplifies the denominator. In this case, let $w=5z$. Then, $dw=5dz$. With this substitution, we have: $$\int \frac{2}{25z^2+25}dz=\int \frac{2}{w^2+25}\cdot \frac{dw}{5}.$$

Integrate

Now, we have: $$\int \frac{2}{w^2+25}\cdot \frac{dw}{5}=\frac{2}{5}\int \frac{1}{w^2+25}dw.$$ This is an arctangent integral, and we know that $\int \frac{1}{u^2+a^2}du=\frac{1}{a}\arctan (\frac{u}{a})+C$, where $C$ is the constant of integration. In our case, $u=w$ and $a=5$, so we have: $$\frac{2}{5}\int \frac{1}{w^2+25}dw=\frac{2}{5}\cdot \frac{1}{5}\arctan \left(\frac{w}{5}\right)+C=\frac{2}{25}\arctan \left(\frac{w}{5}\right)+C.$$

Reverse the substitution and find the final answer

Finally, we reverse the substitution by replacing $w$ with $5z$: $$\int \frac{2}{25z^2+25}dz=\frac{2}{25}\arctan \left(\frac{5z}{5}\right)+C=\frac{2}{25}\arctan (z)+C.$$ This is the indefinite integral of the given function.

Check the answer by differentiation

Now, we need to differentiate the indefinite integral we just found to check if we get the original function. So, let $F(z)=\frac{2}{25}\arctan (z)+C$. Differentiating $F(z)$ with respect to $z$, we get: $$F^{\prime }(z)=\frac{d}{dz}(\frac{2}{25}\arctan (z)+C)=\frac{2}{25}\cdot \frac{1}{1+z^2}.$$ As you can see, $F^{\prime }(z)=\frac{2}{25z^2+25}$, which is the same as the original function. This confirms that our indefinite integral is correct: $$\int \frac{2}{25z^2+25}dz=\frac{2}{25}\arctan (z)+C.$$

Key concepts

Substitution Method

The substitution method is a powerful tool when it comes to solving complicated integrals. It involves changing the variable of integration to simplify the integral into a form that's more manageable. In the given problem, we encounter an integral of the form $\int \frac{2}{25z^2+25}dz$. The presence of the quadratic expression under the denominator suggests that substitution might make the integration easier.

In our case, we noticed that the denominator $25z^2+25$ resembles a perfect square form. By introducing a new variable, we can simplify the expression. We chose$w=5z$, leading to $dw=5dz$. This substitution transforms our integral to $\int \frac{2}{w^2+25}\cdot \frac{dw}{5}$, thus paving the way for easier integration. This method effectively converts the integral into an arctangent form, which is much simpler to handle in this context.

By using substitution, we often transform a complex integral into a standard form that is easier to integrate. This technique not only simplifies the integration process but also helps in identifying integral forms that match known formulas.

Arctangent Integral

Arctangent integrals occur frequently when integrating rational functions with quadratics in the denominator. The integral $\int \frac{1}{u^2+a^2}du$ is a standard form that results in the arctangent function $\frac{1}{a}\arctan \left(\frac{u}{a}\right)+C$, with $C$ as the constant of integration.

In the simplified version of our indefinite integral $\frac{2}{5}\int \frac{1}{w^2+25}dw$, we easily identified it as an arctangent integral because the expression $w^2+25$ fits the form $u^2+a^2$ where $a=5$.

This recognition allowed for a straightforward integration resulting in the solution: $\frac{2}{25}\arctan \left(\frac{w}{5}\right)+C$. It demonstrates how recognizing integral forms can lead directly to solving an otherwise complex problem without any trial and error.

Differentiation Check

A differentiation check is the final step in verifying the correctness of an indefinite integral solution. Once an integral is evaluated, differentiating the result should return the original integrand.

For our integral, we found that the antiderivative was $F(z)=\frac{2}{25}\arctan (z)+C$. Differentiating $F(z)$ with respect to $z$ involves using the derivative $\frac{d}{dz}\arctan (z)=\frac{1}{1+z^2}$. Therefore, the derivative of $F(z)$ is $F^{\prime }(z)=\frac{2}{25}\cdot \frac{1}{1+z^2}=\frac{2}{25z^2+25}$.

The differentiation correctly returns us to the original function $\frac{2}{25z^2+25}$. This step reinforces our confidence that the computed integral $\frac{2}{25}\arctan (z)+C$ is indeed the correct solution of the problem.