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Trying to evaluate this indefinite integral:

$ \int (x^2 + 1)\cos2xdx$

So far I have the following: $u=x^2 + 1 \Rightarrow du = 2xdx$ and $dv=\cos2x \Rightarrow v = \frac {\sin2x}{2}$. So the integral is equal to:

$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\int {\frac{\sin2x}{2}}2xdx$

Next, I make another substitution for the integral on the right hand side; let $ u = x \Rightarrow du = dx$ and let $dv = \sin2x \Rightarrow v = \frac {-\cos2x}{2}$. Now I have the following:

$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left (-\frac {x\cos2x}{2} - \int -\frac {cos2x}{2}dx\right)$

Which after integrating becomes:

$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left(-\frac {x\cos2x}{4} + \frac {\sin2x}{4}\right)$

But when solving with the integrator on my calculator, I get a different answer (it looks like I am getting closer, but still off). What am I doing wrong here??

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    @Michael: Yes, sir.2011-12-06

1 Answers 1

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I see three mistakes in your calculations:

  1. As Andre pointed out in the comments, in the first substitution $du = 2x dx$.
  2. There is a sign error in the second integration by parts: $-\left(-\frac{x\cos 2x}{2} - \int \frac{-\cos 2x}{2} dx\right) = \frac{x\cos 2x}{2} - \int \frac{\cos 2x}{2} dx.$
  3. An integration constant should appear as early as the first integration by parts.
  • 0
    Ah, ok. The sign error fixed it - I should have put parentheses around the second integration by parts to avoid that error. And I added the integration constant. Thanks.2011-12-05