Evaluate the definite integral. ∫ 0 π t 2 sin t d t \int_0^{\pi}t^2\sin tdt

this problem, we are asked to evaluate the definite integral with the integral of t squared times the sine of t, and that's going to be bounded from zero to pi. So how do I go about solving this problem? Well, what I need to do is take a look at the function we're integrating here, and notice we have a product of two functions. Typically, when I see this, the first thing I'm gonna try to do is a variable substitution. So I might try doing a variable substitution on the t squared there. Now if I say u to t squared, the derivative would be two t. But notice that the two t that we get from the derivative is not going to get rid of anything that's inside the sine function, so that's not gonna work. Now I could set this t equal to u inside the sine function, but then the derivative of t with respect to t would just be one. And that's not going to allow us to clear anything with this t squared out front here. So variable substitution is not going to work. So what we need to do is use integration by parts. So when it comes to integration by parts, we first need to find what's going to be our u. Well, looking at this, u is gonna be whatever becomes more simple when I take its derivative. And And I can see that we have this t squared that shows up right here. Now if I were to take the derivative of sine, what I'd end up getting is cosine. Then I'd get the derivative of cosine, which would be negative sine, and just it would we would just continue to go back and forth between sine and cosine, and that's just not gonna help us. So t is definitely what I wanna set equal to u here. So we're gonna have u equals t squared, meaning our derivative of u is gonna be the derivative of t squared with respect to t, which is just gonna be two t dt. Now next, I need to find my dv. Dv needs to be whatever is easily integrated here. Now I see that we have this sine of t here, and I know that dealing with these trig functions like sine and cosine, I can actually integrate those in a relatively straightforward way, so I'm going to say that the sine of t is dv. So dv equals the sine of t, and then the integral here with respect to t, well, I just see that's going to give me v, and integrating sine well it's going to give me negative cosine. So that is how we can find our dv and our v and then our u and our du, and now we can plug things into this equation. So writing this all out we're going to have u times v which is going to be t squared times negative cosine t, so I'm going to write this as negative t squared cosine of t. Then we're going to have minus the integral, and then it's going to be of v du. So multiplying these together we're going to have negative cos t, and we're going to have du which is two t dt. Now what do I do from here? Well at this point, notice what I need to do is apply bounds because we have that this goes from zero to pi. So you need to apply the bounds on this portion right here, which will go from zero to pi, then you need to also put bounds on this integral, which goes from zero to pi, same bounds. Now I can try to make this a little bit more simple by canceling these negative signs and pulling this two out front. This is gonna give me negative t squared cosine of t. That's gonna go from zero to pi, and then we're going to have plus, and then because we cancel these negative signs here, so it's gonna be plus. It's gonna be the integral from zero to pi. Then we're gonna have two on the outside, and then integrate cosine t times t with respect to t. Now notice something that happens here. This is not something that we can integrate. We can try to do a variable substitution here, but since we have just a t there and just a t there, a variable substitution is not gonna work. So we need to use integration by parts a second time. So I'm actually going to wait to apply these bounds from zero to pi and first do this integration by parts process again. So let's go ahead and try this. So I need to find another u that I set. Now last time we set the u equal to t squared, and that seems to get things to be more simple. We went from a t squared in the original integral to just a t. So I'm going to take just that t instead of equal to u, meaning our derivative of t is just going to be one dt. Now next I need to find my dv, and what I'm going to do from here is choose this cosine term right here. So I'm going to say that dv is going to be the cosine of t, and if I go ahead and integrate this, well that's gonna give me v, and integrating cosine is just going to give me the sine of t. So what I can do is write this all out again, where we replace this part with our new integration by parts strategy that we're applying once again. So writing this all out, we're going to have negative t squared cosine of t. That's bounded from zero to pi. They're going to have plus two times, and then what we need to go ahead and do is use this equation again. So we're going to have u times v, which in this case is going to be t times the sin of t, so we're going to have t sine of t, and then it's going to be bounded right here from zero to pi, and we're going to have minus the integral that goes from zero to pi of v du, which is going to be the sine of t integrated with respect to t. So this is what everything comes out to for our integral. Now as you can see it's important to keep track of these bounds because we will need to apply these at the end here, but for now I'm still going to focus on just simplifying this integral and seeing if I can get to a final result. Now we'll have this negative t squared cosine of t out front. That's going to be from zero to pi. Then we have plus and then this two here times t times the sine of t from zero to pi. And what I know is that the integral of sine is going to be negative cosine, and that negative cosine will cancel with a negative right there, so we're going to get plus and that's going to be cosine of t from zero to pi. Then we're gonna have, and we actually don't have to write the cost of integration c. I was just about to say that, but we don't have to write it because it's definite integral. So this right here is where we're going to finish off. So now that we have everything figured out, we can recognize that all of this, this entire thing that we've just derived, and I'm gonna go ahead and simplify this by distributing the two into the brackets here. So now that we've solved for all this, we're going to get negative t squared cosine of t. That's going to be plus two t sine of t. That's going to be plus two cosine of t. And all of this, this entire thing is going to be bounded from zero to pi. So notice it really is just the same process that we've always used with integration by parts. We just have to apply bounds at the very end. So to do this, well, what I'm first gonna do is plug in my high bound of pi, so my upper bound. And so plugging in pi for every place that I c t, well, that's going to give me negative pi squared times the cosine of pi plus, and then we're going to have two times pi, that's going to be times the sine of pi plus, and then we're going to have two times the cosine of pi. That's all of that. Then I need to subtract off my lower bound plug then. So my lower bound here is zero. So what we're gonna have is negative zero squared, and and notice here that the zero is just gonna set all of that to zero. The zero here is gonna set all of this to zero. So all we're gonna have is just two cosine of zero. And I know the cosine of zero is just one. This is gonna be two times one, which is just two. So we end up having is all of this that we have right over here minus two. So this is something that you can go ahead and plug into your calculator to see what result you get. And as a decimal approximation, it should come out to about 5.87 if you plug this all in exactly as you see it. So if you want just the raw answer, it's just going to be everything that we have right up here, this value that we figured out. But if you want the decimal approximation, this should be about 5.87. So this is how you can solve these problems, but you have to use repeated integration by parts, and you also have bounds that you need to apply at the end or basically just a definite integral. See you in the next one.

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Evaluate the definite integral.
$\int_{0}^{π} t^{2} \sin t d t$

$7.87$

$5.87$

$negative 9.51$

$negative 9.87$

Verified step by step guidance

Step 1: Recognize that the problem involves evaluating a definite integral of the form $ \int_0^{\pi} t^2 \sin t \, dt $. This requires integration by parts, as the integrand is a product of $ t^2 $ (a polynomial) and $ \sin t $ (a trigonometric function).

Step 2: Recall the formula for integration by parts: $ \int u \, dv = uv - \int v \, du $. Choose $ u = t^2 $ (since it simplifies upon differentiation) and $ dv = \sin t \, dt $ (since it is straightforward to integrate).

Step 3: Compute $ du $ and $ v $: Differentiate $ u = t^2 $ to get $ du = 2t \, dt $. Integrate $ dv = \sin t \, dt $ to get $ v = -\cos t $.

Step 4: Substitute into the integration by parts formula: $ \int_0^{\pi} t^2 \sin t \, dt = \big[ -t^2 \cos t \big]_0^{\pi} + \int_0^{\pi} 2t \cos t \, dt $. Evaluate the first term $ \big[ -t^2 \cos t \big]_0^{\pi} $ and simplify.

Step 5: For the remaining integral $ \int_0^{\pi} 2t \cos t \, dt $, apply integration by parts again. Let $ u = 2t $ and $ dv = \cos t \, dt $. Follow the same process to compute $ du $, $ v $, and substitute into the formula. Combine all terms and evaluate the definite integral.