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JH
Numerade Educator

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Problem 10 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^\pi \sin^2 t \cos^4 t dt $

Answer

$\int_{0}^{\pi} \sin ^{2} t \cos ^{4} t d t=\frac{\pi}{16}$

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Video Transcript

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