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Chapter 5: Q. 88 (page 419)

Prove, in the following two ways, that the signed area under the graph of the function f(x)=sinxcos2xon an interval [-a,a]centered about the origin is always zero:

(a) by calculating a definite integral;

(b) by considering the symmetry of the graph of the functionf(x)=sinxcos2x

Short Answer

Expert verified

Hence proved.

Step by step solution

01

Part(a) Step 1. Given information.

The given function and interval isf(x)=sinxcos2xand[-a,a].

02

Part (a) Step 2. Explanation.

Using definite integral,

鈭�-aasinxcos2xdx=-13cos3(x)-aa=-13cos3(a)-13cos3(-a)=13cos3(a)-13cos3(a)=0

03

Part (b) Step 1. Explanation.

The graph of the function is ,

Now, from the graph, the function is general sine and cosine function with period whose graph is half above and half below the -axis through origin.

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(c) True or False: The substitution x = 2 tan u is a suitable choice for solving鈭�1x2+4dx.

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving鈭�x2+4鈭�5/2dx

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form x2鈭�a2.

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(g) True or False: When using trigonometric substitution with x=asinu, we must consider the cases x>a and x<-a separately.

(h) True or False: When using trigonometric substitution with x=asecu, we must consider the cases x>a and x<-a separately.
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