# Do different methods of integration yields different results?

I have to solve $$\int \sin^3x\cos^3x\,dx$$

I tried substitutions and got the answer $$\frac{-1}{64}(3\cos(2x)-\frac{1}{3}\cos(6x))+C$$

The solution given in book is $$\frac{1}{6}\cos^6x-\frac{1}{4}\cos^4x+C$$

I don't know whether both are equal or not

So I have some questions:

Are both answers same? If no , then which one is correct?

Do different methods of integration results differently? Or are they same , it is just matter of rearrangement?

I am having this problem in many question because I use different method from that mentioned in book, i.e., in substitutions, I reach the result but substituting different thing from that mentioned in solution which leads to different answer. Do different methods of integration produce different results?

• What did you sub? – K Split X Jul 23 '17 at 15:04
• those two expression can at most differ by constant value. So the constant $C$ may be different in those two expression. – MAN-MADE Jul 23 '17 at 15:05
• Similar problem: math.stackexchange.com/q/1659329/9464 – Jack Jul 23 '17 at 15:08
• Shouldn't it be the following? Perhaps a typo? $$\int \sin^3{x}\cos^3{x}~dx=\frac{-1}{64}(3\cos(2x)\color{red}{-}\frac{1}{3}\cos(6x))+C$$ If this is indeed a typo, the arbitrary constants of integration differ by $\frac{1}{24}$, implying that both your answer and the book's answer are correct. – projectilemotion Jul 23 '17 at 15:14

Your answer is almost exactly the answer of Mathematica $\frac{1}{192} \cos (6 x)-\frac{3}{64} \cos (2 x)$

The answer of the book is also correct after simplifying the previous one.

Another way to get the answer of the book is to transform a little the integral

$$\int \sin^3x \left(\cos^2 x\right) \cos x\, dx=\int \sin^3x \left(1-\sin^2 x\right) \cos x\, dx$$ which gives $$\int \left(\sin ^3x -\sin^5 x\right)\, \text{d}(\sin x)=\frac{\sin^4 x}{4}-\dfrac{\sin^6 x}{6}+C$$ which is almost the same result of your book

• After reading your answer I don't know is the book right, is the OP right , are both results correct or none. All results are almost the same. What does this mean? That there is only a "small" error in the derivation of the results? Can different methods produce different results? How can they differ? Why do they differ? How can one check which is correct? How do you get your Mathematica result? – miracle173 Jul 25 '17 at 10:22
• Sin and cos are linked by the identity $\sin^2 x+\cos^2 x=1$ therefore you can transform my result in the other one by replacing $\sin^2 x$ with $1-\cos^2 x$ They only look different like in $2+2=1+3=0+4$ – Raffaele Jul 25 '17 at 13:17

As I told in comment:

Let $F(x)$ and $G(x)$ are two function such that $F'(x)=G'(x)=f(x)$, then $F(x)-G(x)$ is constant(Why?). Then $\int f(x)dx=F(x)+C_1=G(x)+C_2$, $C_1,C_2$ are constant(may be same sometimes).

Then if your answer is not same as the text book answer then Your Answer-Textbook Answer=constant. So you can reach to your textbook answer by simplifying it. Note that $C$ in these to expression may be different then!

P.S. The answer you got is wrong. See this. Note that red curve and blue curve are the answer you could get. But you got green curve, which is wrong. Also note that red curve and blue curve same but red curve slide upward a little (constant difference!).

You can check their answer is the same as yours with De Moivre's formula.

To have lighter notations, we'll set $u=\mathrm e^{ix}$, $\bar u=\mathrm e^{-ix}$, and use the rules of computation: $$u\mkern2mu\bar u=1,\quad u^n+\bar u^n=2\cos nx.$$

Let's linearise their answer: \begin{align} \frac{1}{6}\cos^6x-\frac{1}{4}\cos^4x&=\frac{1}{6}\Bigl(\frac{u+\bar u}2\Bigr)^6-\frac{1}{4}\Bigl(\frac{u+\bar u}2\Bigr)^4&&(384=6\cdot 64)\\ &=\begin{aligned}[t]\frac{1}{384}(u^6+6&u^4+15u^2+15\bar u^2+6\bar u^4 +\bar u^6)\\ -\frac1{64}(&u^4+4u^2+6+4\bar u^2+\bar u^4) \end{aligned}\\ &=\frac{1}{384}\bigl(u^6+\bar u^6-9(u^2+\bar u^2+4)\bigr)\\ &=\frac1{192}(\cos6x-9\cos 2x-18)\\ &=\frac1{64}\Bigl(\frac13\cos 6x-3\cos 2x-6\Bigl) \end{align} so their answer differs from yours only by a constant and the sign before $\cos 6x$.

• The answer given in OP is in fact not what you obtain, a sign is flipped, as mentioned in a comment. – quid Jul 24 '17 at 12:52
• @quid: Oh! yes, I didn't notice (the O.P. had really small parentheses, for my excuse). I'll mention that in my answer. – Bernard Jul 24 '17 at 13:36

Only the answer $$-\frac{1}{64}\left(3\cos \left(2x\right)\right)+\frac{1}{3}\cos \left(6x\right)$$ is correct.

Look at the following graph: We want the derivative of both the functions you have given to line up with the red graph. As you can see from this image, only the green graph (the $1/6$) one does. I made it so that it only displays values $x>0$ so you can actually see that if it wasn't that, then it would completely overlap.

So your book is right. As for your solution, I can't tell. Please show the steps you took in your solution. I just know that your solution is wrong.

• Oops you have misread , the answer mentioned by you is mine and another one is of book , alternatively you are saying book is wrong@KSplitX – Atul Mishra Jul 23 '17 at 15:18
• You said, "the solution given in the book is .. 1/6 cos^6x ... " Whatever the case is, the 1/6 answer is correct, the other is not – K Split X Jul 23 '17 at 15:21
• But first line of your answer says mine is correct – Atul Mishra Jul 23 '17 at 15:23
• At the start you say the $1/64$ answer is the correct one, then you say the $1/6$ is the correct one. I think this is what confuses @AtulMishra. – quid Jul 24 '17 at 12:50
• The very start of your answer is neither correct, nor one of the answers given in the question. – Thomas Andrews Jul 25 '17 at 16:28

Without seeing your work, I can't figure out where you went wrong, but the correct answer (in your form) is:

$$\frac{-1}{64}\left(3\cos 2x-\frac{1}{3}\cos 6 x\right)+C\tag{1}$$

You can then exand this formula using $\cos 2x=2\cos^2 x-1$ and $$\cos 6x = 32\cos^6x -48\cos^4x+18\cos x-1$$

Substituting these gives the result from the book.

The book's result can be arrived at by noting that:

$$\sin^3 x\cos^3 x = \sin x(1-\cos^2 x)\cos^3 x$$

Then letting $u=\cos x$ and thus $du=-\sin x$ so you are solving:

$$\int (u^2-1)u^3\,du$$

A third answer can be had be noting that $\sin x\cos x=\frac{1}{2}\sin 2x$ so your integral is:

$$\frac{1}{8}\int \sin^32x\,dx=\frac{1}{8}\int(1-\cos^2 2x) \sin 2x\,dx$$

Letting $u=\cos 2x$ then $\sin 2x\,dx = -\frac{1}{2}\,du$ and the integral is:

$$-\frac{1}{16}\int (1-u^2)\,du=-\frac{1}{16}\left(u-\frac{u^3}{3}\right)+C$$

Yielding $$\frac{1}{16}\left(\frac{\cos^3 2x}{3}-\cos 2x\right)+C$$

$$\int \sin^3x\cos^3x.dx=\int \sin^2x\cos^3x.\sin xdx=\\ \int (1-\cos^2x)\cos^3x.\sin xdx$$now take $u=\cos x \to du=-\sin x$ $$\int \sin^2x\cos^3x.sin xdx=\int(1-u^2)u^3.(-du)=\\ \int(u^5-u^3)du=\frac {u^6}{6}-\frac{u^4}{4}\\=\frac {(\cos x )^6}{6}-\frac{(\cos x )^4}{4}+c$$ or $$\sin^3x\cos^3x=(\frac{2}{2}\sin x\cos x)^3=\\\frac18(\sin(2x))^3$$ you know $\sin(6x)=3\sin(2x)-4\sin^3 (2x)$ $$\frac18(\sin(2x))^3=\frac18(\frac14(3\sin(2x)-\sin(6x)))\to\\ \int \sin^3x\cos^3x.dx=\int \frac{1}{32}((3\sin(2x)-\sin(6x))dx=...$$