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I attempted this problem, but when I tried taking the derivative of my answer to check my work, it did not seem to match up. I cannot figure out where I went wrong. Any ideas? These are my steps - I used half angle identities: $$\displaystyle I=\int\cos^4x\sin^6x\ dx$$ using the identities: $\displaystyle \sin x\cos x=\frac{\sin(2x)}{2}$ and $\displaystyle \sin^2x=\frac{1-\cos(2x)}{2}$: $$\displaystyle \int\left(\frac{\sin(2x)}{2}\right)^4\left(\frac{1-\cos(2x)}{2}\right)\ dx$$ $$\displaystyle \frac{1}{32}\int(\sin^4(2x))(1-\cos(2x))\ dx = \frac{1}{32}\int\sin^4(2x)-(\cos(2x))(\sin^4(2x))\ dx$$ We can now divide this integral into two parts: $$\displaystyle \frac{1}{32}\int\sin^4(2x)\ dx$$ and $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$ Let's integrate the first part first: $$\displaystyle \frac{1}{32}\int\sin^4(2x)\ dx$$ $$\displaystyle =\frac{1}{32}\int(1-\cos^2(2x))(\sin^2(2x)\ dx$$ $$\displaystyle =\frac{1}{32}\int(\sin^2(2x)-\sin^2(2x)(\cos^2(2x))\ dx$$ $$\displaystyle =\frac{1}{32}\int\frac{1-\cos(4x)}{2}-\frac{\sin^2(4x)}{4}\ dx$$ $$\displaystyle =\frac{1}{128}\int2-2\cos(4x)-\sin^2(4x)\ dx$$ $$\displaystyle =\frac{1}{128}\int2-2\cos(4x)-\frac{1-\cos(8x)}{2}\ dx$$ $$\displaystyle =\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{1}{256}\int1-\cos(8x)\ dx$$ $$\displaystyle =\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}+C$$ Now, the second part: $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$ Let's do a u-substitution $$\displaystyle u=\sin(2x)\ \ \ \ \ \ \ \ \frac{du}{dx}=2\cos(2x)\ \ \ \ \ \ \ \ \frac12du=\cos(2x)dx$$ Now we have: $$\displaystyle \frac{-1}{64}\int u^4\ du$$ $$\displaystyle \frac{-u^5}{620} = \frac{-\sin^5(2x)}{620}+C$$ Now, putting everything together: $$\displaystyle I=\frac{x}{64}-\frac{\sin^4(2x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{620}+C$$

$$\displaystyle =\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{620}+C$$

Correction given from answers:

Now, the second part: $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$ Let's do a u-substitution $$\displaystyle u=\sin(2x)\ \ \ \ \ \ \ \ \frac{du}{dx}=2\cos(2x)\ \ \ \ \ \ \ \ \frac12du=\cos(2x)dx$$ Now we have: $$\displaystyle \frac{-1}{64}\int u^4\ du$$ $$\displaystyle \frac{-u^5}{320} = \frac{-\sin^5(2x)}{320}+C$$ Now, putting everything together: $$\displaystyle I=\frac{x}{64}-\frac{\sin^4(2x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C$$

$$\displaystyle =\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C$$

Burt
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2 Answers2

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Everything is correct except for some simplification right there at the bottom when you do $u^4$. Anti-derivative of $u^4$ is $u^5/5$. So,the denominator is 320 not 620.

SL_MathGuy
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  • Checking in Wolfram alpha indicate that this is the only error. – Arctic Char Dec 01 '19 at 06:46
  • But, doesn't the sin(2x) make it 320 times two? – Burt Dec 01 '19 at 06:56
  • oh, I see. Since it is part of the substitution I'm not actually integrating that part. – Burt Dec 01 '19 at 07:01
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    320 comes from 5*64. 5 comes from $u^5/5$. So, when you evaluate the second integral in terms of $u$, you get $\frac{-u^5}{320} + constant$. Then, plug $u=sin(2x)$ back in. You already considered 2 in $sin(2x)$ when you do the integration. No need to reconsider it. – SL_MathGuy Dec 01 '19 at 07:02
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You did a good job (except the minor error) but it could have been simpler to expand first in terms of cosines to get

$$A=\cos^4(x)\sin^6(x)=-\cos^{10}(x)+3 \cos^8(x)-3 \cos^6(x)+\cos^4(4)$$

Now, using the Power-reduction formula, we can interchange between $\cos^n(x)$ and $\cos(nx)$ as (for even $n$)

$$\cos^n(x) = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} \binom{n}{k} \cos{((n-2k)x)}$$

Applying it, this should give $$A=\frac{3}{256}-\frac{1}{256} \cos (2 x)-\frac{1}{64} \cos (4 x)+\frac{3}{512} \cos (6 x)+\frac{1}{256} \cos (8 x)-\frac{1}{512} \cos (10 x)$$ and then the simple antiderivative $$I=\int\cos^4(x)\sin^6(x)\, dx=\frac{3 x}{256}-\frac{\sin (2 x)}{512} -\frac{\sin (4 x)}{256} +\frac{\sin (6 x)}{1024}+\frac{\sin (8 x)}{2048}-\frac{\sin (10 x)}{5120}$$

Claude Leibovici
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  • That does seem simpler. I do not know the Power-Reduction formula. At what level of mathematics is it learned? – Burt Dec 01 '19 at 15:02
  • @Burt. You know the multiple angles formulae going from $\cos(nx)$ to $\cos^n(x)$. Then $\cos^n(x)$ can be expressed as a linear combination of $cos(kx)$. – Claude Leibovici Dec 01 '19 at 15:07