Showing that integral is an induction

In summary, the conversation discusses using the recursive method to prove a given equation, which involves substituting n with n+2 and repeatedly applying the substitution until reaching the desired result. This method is found to be more interesting than using proof by induction. At the end, the resulting integral is equal to 1.
  • #1
jisbon
476
30
Homework Statement
Show that ##\int _0^{\frac{\pi }{2}}\:sin^{2n+1}x\:dx = \frac{2*4*6*...2n}{3*5*7*...(2n+1)}###
Relevant Equations
##\int _0^{\frac{\pi }{2}}\:sin^{n}x\:dx\:=\:\frac{n-1}{n}\int _0^{\frac{\pi }{2}}\:sin^{n-2}x\:dx##
Hi all,

Having this equation derived:
##\int _0^{\frac{\pi }{2}}\:sin^{n}x\:dx\:=\:\frac{n-1}{n}\int _0^{\frac{\pi }{2}}\:sin^{n-2}x\:dx##
What I will do is simply substitue n with n+2, and I will get the following:
##\frac{2n}{2n+1}\int_{0}^{\pi /2}(sinx)^{2n-1}dx##
What should I do from here?
 
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  • #2
Do you understand the concept of proof by induction?
 
  • #3
You can either do induction or use recursion.

I found it to be more interesting to follow the recursive way on this.

By applying the equation you give at homework equations where you substitute n with 2n-1 you get something and then you ll apply it again for n=2n-3 and then for n=2n-5 and so on and with this recursive way you ll get to prove the desired result given that at the very end you ll be left with ##\int_0^{\frac{\pi}{2}}\sin x dx## which is equal to 1.
 

Related to Showing that integral is an induction

What is the concept of induction in mathematics?

Induction is a mathematical technique used to prove that a statement is true for all values of a variable. It involves proving that the statement is true for a base case and then showing that if the statement is true for a given value, it is also true for the next value.

How can induction be used to show that an integral is true?

To show that an integral is true using induction, we must first prove that the statement is true for a base case, usually when the variable is equal to 0. Then, we must show that if the integral is true for a given value, it is also true for the next value. This can be done by using the induction hypothesis and manipulating the expression until it matches the expression for the next value.

What are the steps involved in using induction to prove an integral?

The steps for using induction to prove an integral are as follows:
1. Show that the statement is true for the base case.
2. Assume that the statement is true for a given value.
3. Manipulate the expression using the induction hypothesis.
4. Show that the manipulated expression matches the expression for the next value.
5. Therefore, the statement is true for all values of the variable.

What are some common mistakes made when using induction to prove an integral?

Some common mistakes when using induction to prove an integral include:
- Forgetting to prove the base case
- Using the wrong induction hypothesis
- Skipping steps and not showing all the necessary manipulations
- Incorrectly matching the manipulated expression to the expression for the next value
- Not clearly stating the induction hypothesis and the manipulations used
It is important to carefully check each step and make sure all the necessary details are included in the proof.

Can induction be used to prove all integrals?

No, induction can only be used to prove certain types of integrals, specifically those that involve a discrete variable such as summations or products. It cannot be used to prove continuous integrals, as these involve an infinite number of values and cannot be reduced to a finite number of steps. Other methods, such as the fundamental theorem of calculus, must be used to prove continuous integrals.

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