Standard +0.3 This is a straightforward double integration by parts question with standard functions (polynomial times trigonometric). While it requires two applications of integration by parts and careful bookkeeping with the factor of 2 in sin(2x), it follows a completely routine procedure with no conceptual challenges. The definite integral evaluation is also mechanical, making this slightly easier than average.
Integrate by parts and reach \(ax^2 \cos 2x + b \int x \cos 2x \, dx\)
M1*
Obtain \(-\frac{1}{4}x^2 \cos 2x + \int x \cos 2x\) , or equivalent
A1
Complete the integration and obtain \(-\frac{1}{4}x^2 \cos 2x + \frac{1}{4}x \sin 2x + \frac{1}{4} \cos 2x\), or equivalent
A1
Use limits correctly having integrated twice
DM1*
Obtain answer \(\frac{1}{8}(\pi^2 - 4)\), or exact equivalent, with no errors seen
A1
[5]
Integrate by parts and reach $ax^2 \cos 2x + b \int x \cos 2x \, dx$ | M1* |
Obtain $-\frac{1}{4}x^2 \cos 2x + \int x \cos 2x$ , or equivalent | A1 |
Complete the integration and obtain $-\frac{1}{4}x^2 \cos 2x + \frac{1}{4}x \sin 2x + \frac{1}{4} \cos 2x$, or equivalent | A1 |
Use limits correctly having integrated twice | DM1* |
Obtain answer $\frac{1}{8}(\pi^2 - 4)$, or exact equivalent, with no errors seen | A1 | [5]