How Challenging Is This Definite Integral with a Tangent and Pi Power?

In summary, a definite integral challenge is a mathematical problem that involves finding the area under a curve between two specific points. To solve it, you can use the fundamental theorem of calculus or integration techniques such as substitution and integration by parts. Definite integrals have various applications in fields such as physics, engineering, economics, and statistics. There is no one specific method for solving definite integral challenges, and the method used depends on the function and limits of integration. To check if your solution is correct, you can use an online calculator, take the derivative, or graph the function.
  • #1
anemone
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Evaluate \(\displaystyle \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}\).
 
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  • #2
Here is my solution:

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)
 
  • #3
MarkFL said:
Here is my solution:

We are given to evaluate:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx\)

Using the property of definite integrals \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) and a co-function identity, we may state:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

Adding the two equations, we obtain:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx\)

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}\)

Hence:

\(\displaystyle I=\frac{\pi}{4}\)

Well done, MarkFL!(Clapping) Your answer is spot on!
 

Related to How Challenging Is This Definite Integral with a Tangent and Pi Power?

1. What is a definite integral challenge?

A definite integral challenge is a mathematical problem that involves finding the area under a curve between two specific points. It is a concept commonly used in calculus and is used to solve a variety of real-world problems.

2. How do I solve a definite integral challenge?

To solve a definite integral challenge, you need to use the fundamental theorem of calculus, which states that the definite integral of a function can be found by evaluating the antiderivative of the function at the upper and lower limits of integration and subtracting the results. You can also use integration techniques such as substitution, integration by parts, or trigonometric substitution.

3. What are the applications of definite integrals?

Definite integrals have various applications in fields such as physics, engineering, economics, and statistics. They can be used to find the area under a curve, calculate displacement, velocity, acceleration, work, and other physical quantities. They are also used in optimization problems and finding the average value of a function.

4. Is there a specific method for solving definite integral challenges?

There is no one specific method for solving definite integral challenges. The method you use depends on the type of function and the limits of integration. You may need to use different techniques, such as substitution or integration by parts, for different problems.

5. How can I check if my solution to a definite integral challenge is correct?

You can check your solution to a definite integral challenge by using an online calculator or by taking the derivative of your answer. If the derivative matches the original function, then your solution is correct. You can also check your solution by graphing the function and visually comparing the area under the curve to the result of your definite integral.

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