If In=∫0^(π)/(4) tan^n x d x n being a positive integer then show that (i) In+In-2=(1)/(n-1) (ii

step 1 Hi, everyone given here, I n is a positive integer. When this prove here this relation, this rel

If In=∫0^(π)/(4) tan^n x d x n being a positive integer then...

Key Concepts

Evaluation of Specific Definite Integrals

This concept pertains to utilizing derived recurrence relations and established bounds to compute exact values of integrals for specific cases. In problems where I? or other specific I? values are required, one applies the recurrence relation sequentially or directly formulates the integrals using standard techniques, leading to precise results.

Inequality Estimation in Integration

Inequality techniques in integration involve establishing upper and lower bounds for an integral by comparing the integrand to functions that are known to be larger or smaller within the interval of integration. This method is particularly useful when exact evaluation is challenging, allowing one to assert estimates such as 1/(2(n+1)) < I? < 1/(2(n-1)) for integrals of trigonometric functions.

Trigonometric Integrals

Integrating trigonometric functions, particularly powers of the tangent function, requires specific techniques such as appropriate substitutions and using trigonometric identities. This knowledge is essential when evaluating definite integrals with limits that are based on the periodic properties of trigonometric functions, especially on intervals like [0, ?/4] where the behavior of functions like tan x is well understood.

Recurrence Relations in Integration

This concept involves expressing an integral with a parameter in terms of integrals with lower parameters, forming a recursion that simplifies the calculation. In the context of integrals involving powers of a function, such as tan? x, a recurrence relation is derived to relate I? and I???, reducing a higher order integral to ones that are easier to handle or already computed.

Transcript

00:01 Hi, everyone given here, i n is a positive integer.

00:06 When this prove here this relation, this relation, then this i -fix need to find out.

00:11 So with that, we can write it here, 10x of the power n -n, this would be given as in equals 0 to 5 over 4, 10x, 10 x to the power n -negative 2 times 10 square -exes, 2x, 2x 2x 2x 2nd squared, negative 1, dx.

00:36 So we just distribute this.

00:39 10x2, 2, 2, 2nd square x2.

00:41 We use substitution here.

00:43 So we get 10x to the par n negative 1.

00:49 We just go with that.

00:54 So let me just simplify it in a way.

00:58 So let's go distributed 0 to 5 over 4, then 10x to the power n negative 2, second 2x 2x 2 2 2 2 5 over 4.

01:11 10x to the power n negative 2 and here and negative 2 this will become here i n negative 2 and this if you use substitution here it's integral we put t as 10x so you'll be x x x x you will get here limits from 0 to 105 or 4 is 1 then it will be t to the power n negative 2 d t so, come to the power n negative 1 over n negative 1.

01:42 The limits from 0 to 1.

01:45 So it might be 1 negative 0.

01:47 That is 1.

01:48 So this is coming up to be, you will get it as equals 1 over n negative 1, n negative 1, n n.

01:57 Here, and negative i and negative 2.

02:01 So the first solution we have i .n positive i n negative 2.

02:07 So we have i .n, positive i n negative 2.

02:11 Equals 1 over and negative 1.

02:14 So this part is broke here, the first part.

02:19 So, second part, we see in 0 to pi over 4 in this interval, we wrote a graph of 10x, this is 5 over 2 here at this point.

02:34 The graph is of 10x you can see.

02:37 This is pi over 4 here we have a value at pi over 4 that is even as 1.