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How do you evaluate the definite integral #int sin(4x)# from #[0,pi/8]#?

Answer 1

Christopher Allers

#1/4#

This is a fairly common integral.

#intsin(ax)dx=-1/acos(ax)# where #a# is any constant.

So,

#int_0^(pi/8)sin(4x)dx=-1/4cos4x|_0^(pi/8)#

Evaluating, we get

#-1/4cos((4pi)/8)-(-1/4cos(0))=-1/4cos(pi/2)+1/4cos(0)=1/4#

As #cos0=1#

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Answer 2

Emily Ammerman

To evaluate the definite integral (\int \sin(4x) , dx) from (0) to (\frac{\pi}{8}):

Use the integral property: [ \int \sin(ax) , dx = -\frac{1}{a} \cos(ax) + C ]

Apply the property with (a = 4):

[ \int \sin(4x) , dx = -\frac{1}{4} \cos(4x) + C ]

Evaluate the integral from (0) to (\frac{\pi}{8}):

[ \left[-\frac{1}{4} \cos(4x)\right]_0^{\frac{\pi}{8}} ]

Substitute the upper and lower limits:

[ -\frac{1}{4} \cos\left(4 \cdot \frac{\pi}{8}\right) - \left(-\frac{1}{4} \cos(0)\right) ]

Simplify:

[ -\frac{1}{4} \cos\left(\frac{\pi}{2}\right) + \frac{1}{4} \cos(0) ]

[ -\frac{1}{4} \cdot 0 + \frac{1}{4} \cdot 1 ]

[ 0 + \frac{1}{4} ]

[ \frac{1}{4} ]

So, the value of the definite integral is (\frac{1}{4}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.