∫0^π sin^5((x)/(2)) is equal to (a) (16)/(15) (b) (15)/(16) π(c) (16)/(15) π^2 (d) (15)/(16).

Name: Problem 7, Chapter 6: Higher Engineering Mathematics
Uploaded: 2022-04-11T02:41:11-08:00
Duration: 9 min 21 s
Channel: Jeff Vermeire
Description: ∫0^π sin^5((x)/(2)) is equal to (a) (16)/(15) (b) (15)/(16) π(c) (16)/(15) π^2 (d) (15)/(16).

∫0^π sin^5((x)/(2)) is equal to (a) (16)/(15) (b) (15)/(16)...

Key Concepts

Definite Integration

This concept refers to the process of evaluating an integral with specified lower and upper limits. In definite integrals, the result is a number that represents the net area under the curve described by the integrand between these boundaries.

Substitution Method

A common technique used to simplify integrals by changing the variable of integration. This is particularly useful when the integrand contains a composite function, as it transforms the integral into a simpler or more standard form that is easier to evaluate.

Integration of Powers of Trigonometric Functions

This technique involves approaches such as power reduction formulas or trigonometric identities to express higher powers of sine or cosine in terms of lower powers or other trigonometric functions. It is essential for evaluating integrals where trigonometric functions are raised to a power.

Trigonometric Identities

These are equations involving trigonometric functions that are true for all values within the domains of the involved functions. They are used to simplify and transform integrals involving trigonometric functions, enabling more straightforward computation of the integral.

Transcript

00:01 We are given the integral sign of x over 2 to the 5th power dx from pi to 0.

00:08 To solve this, we want to make a simple u substitution.

00:12 We'll get u equal x over 2.

00:16 Then du equals d x over 2 which we can rewrite is 2du equals d x now we can rewrite our integral sign the fifth power of u to you can't forget that two it goes out in the front now we need to find our new limits of integration so u when x equals 0 over 2 equals 0 u when x equals pi over 2 equals pi over 2 so we have 0 and pi over 2 so we have 0 pi over 2 next we need to use a reduction formula which is the integral sign to some power m dx equals negative cosine x sine x sine x sine x over oh that's over m plus m minus 1 over m integral sine m minus 2 d x and here m equals 5 so now we just plug that in and i'll change colors here so this equals cosine of u times sine to the fourth power of u over 4 no, that should be 5.

02:43 And i should also have that multiplied by our negative 2.

02:53 And then we add this integral, which is fifth times our 2.

03:06 And then we have the limits of integration, so into the third power u, d .u.

03:19 So now we have to take this first bit.

03:25 At the limits of integration, which equals negative 2, 5th, sine, pi over 2 times a cosine, pi over 2, minus negative 2, 5th, sine of 0.

04:25 So this is negative 2, 5th times 1 times 0 minus negative 2, 5ths times 0 times 1.

04:36 So all of this equals 0.

04:39 So we can ignore this...