A-Level Edexcel Maths Pure Integration: 2. (a) Use integration by parts

2. (a) Use integration by parts to find $\, \int x \sin 3x \, dx.\ (b) Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.\ - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8

Question 4

$2.-(a)-Use-integration-by-parts-to-find--$\,-\int-x-\sin-3x-\,-dx.\----(b)-Using-your-answer-to-part-(a),-find--$\,-\int-x^2-\cos-3x-\,-dx.\-Edexcel-A-Level Maths Pure-Question 4-2012-Paper 8.png$

2. (a) Use integration by parts to find $\, \int x \sin 3x \, dx.\ (b) Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.\

Worked Solution & Example Answer:2. (a) Use integration by parts to find $\, \int x \sin 3x \, dx.\ (b) Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.\ - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8

Step 1

Use integration by parts to find $\, \int x \sin 3x \, dx.$

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Answer

To solve $\, \int x \sin 3x \, dx,$ we use the integration by parts formula:

$\int u \, dv = u v - \int v \, du$

Here, we can choose:

Let $u = x$ , so $du = dx$
Let $dv = \sin 3x \, dx$ , so $v = -\frac{1}{3} \cos 3x$

Substituting these into the formula, we get:

$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x - \int -\frac{1}{3} \cos 3x \, dx$

Now we evaluate the integral:

$\int \cos 3x \, dx = \frac{1}{3} \sin 3x + C$

So, substituting back, the expression becomes:

$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{3} \cdot \frac{1}{3} \sin 3x + C$

Thus, simplifying gives:

$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C$

Step 2

Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.$

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Answer

Using the result from part (a), we apply integration by parts again for $\int x^2 \cos 3x \, dx$ .

Let:

$u = x^2$ so $du = 2x \, dx$
$dv = \cos 3x \, dx$ , so $v = \frac{1}{3} \sin 3x$

Applying integration by parts gives:

$\int x^2 \cos 3x \, dx = \frac{1}{3} x^2 \sin 3x - \int \frac{2}{3} x \sin 3x \, dx$

From part (a), we know: $\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C$

Substituting this back into our expression gives:

$\int x^2 \cos 3x \, dx = \frac{1}{3} x^2 \sin 3x + \frac{2}{3} \left( -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x \right) + C$

Thus, simplifying further yields: $\int x^2 \cos 3x \, dx = \frac{1}{3} x^2 \sin 3x - \frac{2}{9} x \cos 3x + \frac{2}{27} \sin 3x + C$

2. (a) Use integration by parts to find $\, \int x \sin 3x \, dx.\ (b) Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.\ - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8

Worked Solution & Example Answer:2. (a) Use integration by parts to find $\, \int x \sin 3x \, dx.\ (b) Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.\ - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 8

Use integration by parts to find $\, \int x \sin 3x \, dx.$

Using your answer to part (a), find $\, \int x^2 \cos 3x \, dx.$

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