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??

  • 5
    $\begingroup$ $du=2x\,dx$, not $du=x$. $\endgroup$ – André Nicolas Dec 5 '11 at 6:48
  • $\begingroup$ Without going through your calculations: did you check if the derivative of your result and the derivative of the calculator's result agree with your original integrand? $\endgroup$ – J. M. is a poor mathematician Dec 5 '11 at 6:49
  • $\begingroup$ @Andre: Arg. I see that mistake now. Forgot to bring the exponent down from $x^2$ and add $dx$. $\endgroup$ – dtg Dec 5 '11 at 6:52
  • $\begingroup$ @J.M. they did not agree. I tried again after taking note of the error which Andre pointed out, but this time I am just slightly off... more above. $\endgroup$ – dtg Dec 5 '11 at 7:04
  • $\begingroup$ "Solve" is the wrong word. You're trying to evaluate the integral. "Solutions" are sought for problems and for equations; values are sought for expressions (including integrals). $\endgroup$ – Michael Hardy Dec 5 '11 at 17:54

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.
  • $\begingroup$ 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. $\endgroup$ – dtg Dec 5 '11 at 7:17

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