WEBVTT
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this problem is from Chapter seven, Section two.
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Problem number ten in the book Calculus Early Transcendence ALS
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eighth Edition by James Store Here we have a definite
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integral from zero a pie sine squared of tea cozy
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into the fourth of TT. Now, since the
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Power's on Sign and co signer both even it will
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be useful to apply these trig identities for sine squared
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and coastlines where so supply these so we can rewrite
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the general bye. So sine squared becomes one minus
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close enough to tea number two and for close into
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the fourth Power will use the fact that coincide to
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the fourth power of tea is co sign square of
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t squared. So using this identity, we can
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write this as one plus co sign two teams all
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over too square. So this is about using this
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fact. So we have one plus co sign two
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tea over to and the whole thing is clear for
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the next step. Let's pull out this constant,
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so have a one half from the first fraction.
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And after we square, we have a one over
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force from the second. So let's multiply those to
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pull out a one over eight Fiona Pie and let's
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also square this numerator, so have a one minus
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co sign. Two teams times one plus two course
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I Tootie plus cosign Square of to Feed DD for
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the next step. Let's just do the symbols application
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here So let's multiply the one through first. So
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we are one plus to co sign Tootie Plus Co
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sign Squared to Team then Well, we'Ll supply everything
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in the second prophecies by negative coz I'm sooty So
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we have a negative co scientist he minus two coastline
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square to tea and then finally, um, negative
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coastline cubed Tootie Titi separated from her scratch. So
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before we integrate, let's combine some like terms So
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here we have in a roll see on a pyre
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one and what can we combine? We have ah
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to co scientist in tea and then a minus coast
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identity So we're left over with cosign Tootie Further,
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we have ah co sign squared to tea and then
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a negative to co sign squared to tea. So
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that's improvised to minus coastlines where Tootie and further we
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still have this negative coast. Thank you. That
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doesn't cancel out with anything. And since that's Ah
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, our power on the coastline. And it's a
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power larger than to what's actually separate that in a
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girl from the original problem, so have a minus
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one over eight integral zero pie cosign cute titi.
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So for this first interval before we evaluate notice here
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we have a co sign square so we could apply
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. Wanna write identities from the very beginning? Two
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toe. Replace that call sign square so we have
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won over eight in a girl's ear. Tobias one
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plus Cose Identity minus. And then now we can
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write this as one half plus coastline of two times
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to tea. So we have a forty now over
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to you and for this in general. Since it's
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odd power of co sign, let's pull out one
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factor of coastline to obtain co sign squared of two
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teams Times Co senti for the next step in the
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first general, it simplifies bunches. We can really
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all here we can do is combine the one in
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the one half, so there's not much to simplify
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here. Cho San to T doesn't simplify co sign
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a forty over two dozen simplify or cancel. So
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let's integrate in girl. One half will become.
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Is he over too? In a world, of
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course identity. If you could do use up here
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, you equals two tea You'LL obtain sign of Tootie
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over and for the General Negative Co sign of Foresee
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Over too. Using another use of U equals forty
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. Yeah, sign of forty over four times two
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, which is eight. It could well be evaluating
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this expression at the values Cirone Pie. For the
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second inaugural, we could set ourselves up for use
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up by rewriting this co sign squared as one minus
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, I swear. Now we plug in the end
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points for the first integral that one over eight plugging
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in pie for tea pi over too sign of two
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by over to minus sign of four pi obree.
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And when we plug in zero first he will have
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zero over too. Signs here over to minus sign
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zero over eight for the second and rule we can
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do a U substitution. U equals a sign of
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beauty So that do you over too? His co
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sign of two teams DT and also since we're doing
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the use of our limits of integration will change.
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So before the end point was zero. The first
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end point was zero. So now it's sign of
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two time zero, which is zero and the other
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end point, a sign of two times pi,
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which is also zero. So the second Integral becomes
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so This can be evaluated using the power rule.
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But here this is one of the properties of the
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definite integral, the integral from a t A zero
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here, able zero. So this integral itself a
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zero. So we don't have to worry about the
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second integral going back to the first Integral, we
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could cancel out sign of two by zero. So
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we was this term sign a four by zero.
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And all these terms are zero. So we're left
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with one over eight times pi over too pi over
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sixteen, and that's your final answer.