积分∫4sin⁴x dx计算错误排查求助

Hey there! Let's walk through your work step by step to spot that sneaky mistake—totally get wanting to figure out where you went off track instead of just switching methods. 😊

I know there is something wrong with my work below, but I'm not sure what. I believe my answer is missing a (3/2)x, but after going through my work three or four times I cannot find the mistake for the life of me. (Yes, I know there is an alternate way of solving the integral, but I'm more interested on finding what I did wrong than finding the correct solution at this point). Thanks for the help.

\begin{align*}
& \int 4\sin^4x , dx & \
&= \int 4\sin^{2x\left(1-\cos}2x \right) dx & \
&= \int \left( 4\sin^2x - 4\sin^2x\cos2x \right) dx & \
&= \int \left( 4\sin^2x - \sin^22x \right) dx & \
&= -4 \int \frac{\cos(2x) - 1}{2} , dx + 4 \int \frac{\cos(4x) - 1}{2} , dx \
&= -2\left[ 0.5\sin2x - x \right] + 2\left[ 0.25\sin4x - x \right] + C \
&= -\sin2x + 2x + 0.5\sin4x - 2x + C \
&= -\sin2x + 0.5\sin4x + C
\end{align*}

Let's break down each step to catch the error:

• Steps 1-4 are all correct:

  • Rewriting $\sin^4x$ as $\sin^2x(1-\cos2x)$ uses a fundamental trig identity correctly.
  • Expanding to $4\sin^2x - 4\sin^2x\cos2x$ is straightforward algebra.
  • Converting $4\sin^2x\cos2x$ to $\sin^22x$ is perfect—since $\sin2x=2\sin x\cos x$, squaring both sides gives $\sin^22x=4\sin2x\cos^2x$, so this substitution checks out.

• The mistake hits at step 5:
  You correctly split the integral into $\int4\sin^2x dx - \int\sin^22x dx$, but added an extra factor of 4 when rewriting the second integral:

  • For the first term: $\int4\sin^2x dx = -4\int\frac{\cos2x - 1}{2}dx$ is actually okay, because $\sin^2x=\frac{1-\cos2x}{2}=-\frac{\cos2x-1}{2}$, so this substitution is valid.
  • For the second term: $-\int\sin^22x dx$ is where you slipped up. Using the identity $\sin^2\theta=\frac{1-\cos2\theta}{2}$ with $\theta=2x$, we get $\sin^22x=\frac{1-\cos4x}{2}$. That means:
    $$-\int\sin^22x dx = -\int\frac{1-\cos4x}{2}dx = \int\frac{\cos4x - 1}{2}dx$$
    You incorrectly added a 4 in front of this integral, which is why your $x$ terms canceled out later (they shouldn't have!).

• Fixing that step to recover the missing term:
  Correct step 5 should be:
  $$= -4 \int \frac{\cos(2x) - 1}{2} , dx + \int \frac{\cos(4x) - 1}{2} , dx$$
  Carrying through the rest of the calculation:

  1. First integral: $-4*\frac{1}{2}\int(\cos2x-1)dx = -2\left[\frac{1}{2}\sin2x - x\right] = -\sin2x + 2x$
  2. Second integral: $\frac{1}{2}\int(\cos4x-1)dx = \frac{1}{2}\left[\frac{1}{4}\sin4x - x\right] = \frac{1}{8}\sin4x - \frac{1}{2}x$
  3. Adding them together:
     $$-\sin2x + 2x + \frac{1}{8}\sin4x - \frac{1}{2}x + C = -\sin2x + \frac{3}{2}x + \frac{1}{8}\sin4x + C$$

That extra 4 was the sneaky culprit—easy to misplace when juggling trig identities and coefficients. Hope that clears up your confusion!

备注：内容来源于stack exchange，提问作者destroyer000lucky