Question Video: Integration by Parts Mathematics • Higher Education

Use integration by parts to find the exact value of ∫_(0) ^(Ο€/4) π‘₯Β² sin 2π‘₯ 𝑑π‘₯.

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Video Transcript

Use integration by parts to find the exact value of the integral of π‘₯ squared sin two π‘₯ 𝑑π‘₯ between zero and πœ‹ by four.

Integration by parts uses the formula that the integral of 𝑒𝑣 dash is equal to 𝑒𝑣 minus the integral of 𝑣𝑒 dash, where 𝑒 dash is the differential of 𝑒, and 𝑣 dash is the differential of 𝑣. Our first step is to split our initial expression π‘₯ squared sin two π‘₯ into 𝑒 and 𝑣 dash. We will let 𝑒 equal π‘₯ squared and 𝑣 dash equals sin two π‘₯.

To work out an expression for 𝑒 dash, we need to differentiate π‘₯ squared. π‘₯ squared differentiated is two π‘₯. To work out an expression for 𝑣, we need to integrate sin two π‘₯, as integration is the opposite of differentiation. The integral of sin two π‘₯ is equal to negative cos two π‘₯ over two. Multiplying π‘₯ squared by negative cos two π‘₯ over two gives us negative π‘₯ squared cos two π‘₯ over two. Multiplying two π‘₯ by negative cos two π‘₯ over two gives us negative two π‘₯ cos two π‘₯ over two.

As we have two negative signs, this could be simplified to negative π‘₯ squared cos two π‘₯ over two plus the integral of two π‘₯ cos two π‘₯ divided by two. We can also cancel the twos after the integration sign, by dividing the numerator and denominator by two. We now need to try and integrate π‘₯ cos two π‘₯. We can integrate this expression using parts once again.

We will let 𝑒 equal π‘₯ and 𝑣 dash equal cos two π‘₯. Differentiating π‘₯ gives us one. Therefore, 𝑒 dash is equal to one. Integrating cos two π‘₯ gives us sin two π‘₯ over two. Multiplying 𝑒 and 𝑣, π‘₯ and sin two π‘₯ over two, give us π‘₯ sin two π‘₯ over two. Multiplying 𝑒 dash and 𝑣 gives us sin two π‘₯ over two. We are now left with negative π‘₯ squared cos two π‘₯ over two plus π‘₯ sin two π‘₯ over two minus the integral of sin two π‘₯ over two.

Integrating the third term, sin two π‘₯ over two, gives us negative cos two π‘₯ over four. Once again, our two negative sins can turn into a positive. This means that the integral of π‘₯ squared sin two π‘₯ is equal to negative π‘₯ squared cos two π‘₯ over two plus π‘₯ sin two π‘₯ over two plus cos two π‘₯ over four. Our final step is to substitute our two limits, πœ‹ by four and zero, and subtract the two answers.

Firstly, let’s substitute π‘₯ equals πœ‹ by four. Before starting, it’s worth noting that our trigonometrical functions are cos two π‘₯ and sin two π‘₯. Therefore, we need to calculate cos of two πœ‹ by four and sin of two πœ‹ by four. Two πœ‹ by four radians is the same as πœ‹ by two radians. And πœ‹ by two radians is equal to 90 degrees. We know from our trig graphs that cos of 90 is equal to zero. Therefore, cos of πœ‹ by two radians must also equal zero.

The sin of 90 degrees is equal to one. Therefore, the sin of πœ‹ by two radians is also equal to one. As cos of πœ‹ by two is equal to zero, we know that the first and third terms in our expression will be equal to zero when π‘₯ is equal to πœ‹ by four. The only term that will give us a value is π‘₯ sin two π‘₯ over two. As π‘₯ is equal to πœ‹ by four and sin of two π‘₯ is equal to one, this term gives us πœ‹ by four multiplied by one divided by two. This is equal to πœ‹ by eight, as πœ‹ by four divided by two is equal to πœ‹ by eight. When π‘₯ is equal to πœ‹ by four, the integral of π‘₯ squared sin two π‘₯ equals πœ‹ by eight.

We also need to consider the lower limit when π‘₯ is equal to zero. Once again, from our trig graphs, we know that cos of zero is equal to one and sin of zero is equal to zero. This means that the middle term π‘₯ sin two π‘₯ over two will be equal to zero. It is possible that the first and third term will have nonzero values, as cos of two π‘₯ is equal to one.

The first term negative π‘₯ squared cos two π‘₯ over two also has an π‘₯ term. And as π‘₯ is equal to zero, this whole term will also equal zero. Therefore, the only nonzero term is cos of two π‘₯ over four. We already know that cos two π‘₯ is equal to one. Therefore, cos two π‘₯ over four is equal to one-quarter. As we have worked out exact values for the upper and lower limit, we can now say that the integral of π‘₯ squared sin two π‘₯ between zero and πœ‹ by four is πœ‹ by eight minus one-quarter.

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