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I am currently taking a calculus class. My teacher while teaching a specific type of indefinite integral told one mysteriously beautiful of the solving the integral. The general form of the integral was -

$\int \sin^m x \cos^n x dx $

He said when both m and n are odd, for example -

$\int \sin^3x \cos^5xdx$

There are 2 methods to solve this which gave 2 completely different answers and both were correct. Let me elaborate.

Method 1 -

$\int \sin^3x \cos^5xdx = \int \sin^2x \cos^5x \sin xdx$

Now let, $\cos x = u$

$\implies -\sin x dx = du$

$\implies -\int u^5(1-u^2)du$

$=\int \big(u^7 - u^5 \big)du$

$= {u^8\over 8} - {u^6\over 6}+c$

$= {\cos ^8x \over 8} - {\cos ^6x \over 6} + c$

Method 2 -

$\int \sin^3x \cos^5xdx = \int \sin^3x\cos^4x\cos xdx$

Now this time let, $\sin x=v$

$\implies \cos xdx = dv$

$\implies \int v^3\big(1-v^2)^2dv$

$= \int \big( v^7 - 2v^5 +v^3\big)dv$

$= {\sin^8x \over 8}+{\sin^4x \over 4} - {1\over 3}\sin^6 x+c_1$

As said before, my teacher said both these answers are correct. My questions are -

  • How are both the answers correct?
  • How do we interpret this result?
HarshDarji
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    Use a bunch of trig identities and get that one differs from the other by a constant. They are the same answer. – Randall Aug 09 '21 at 13:30
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    Recall... $\sin^2(x)+\cos^2(x)=1$ so wherever you see $\sin^2(x)$ you could have instead written $(1-\cos^2(x))$... Recall also that an indefinite integral of a function is a whole class of functions, each of whom differ by a constant. When taking a definite integral, yes, the constants wind up cancelling out, but for indefinite integrals the "+C" is very much relevant and part of the answer. – JMoravitz Aug 09 '21 at 13:30
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    Think of it this way: if I told you that I had answers $x^2+2x+1$ and $(x+1)^2$, would you really say they are different? Yes, they look different, but.... – Randall Aug 09 '21 at 13:30
  • Ohh okay. So they're the same thing but written in two different ways. Thanks for the help everyone! – HarshDarji Aug 09 '21 at 13:32
  • Use again the identity $cos^2=1-sin^2$ in the first result, for example. – Dog_69 Aug 09 '21 at 13:35

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if you substitute $cos^2\theta=1-sin^2\theta$ into your first expression you get (for tidiness I'm letting $cos\theta=C$ and $sin\theta=S$) $$ \frac{C^8}{8}-\frac{C^6}{6}+c_0$$ $$ =\frac{C^6}{2}(\frac{C^2}{4}-\frac{1}{3})+c_0$$ $$ =\frac{(1-S^2)^3}{2}(\frac{1-S^2}{4}-\frac{1}{3})+c_0$$ $$ =\frac{(1-S^2)^3}{2}(\frac{-1-3S^2}{12})+c_0$$ $$ =-\frac{1}{24}((S^2)^3-3(S^2)^2+3(S^2)-1)(1+3S^2)+c_0$$ $$ =-\frac{1}{24}((S^6)-3(S^4)+3(S^2)-1)(1+3S^2)+c_0$$ $$ =-\frac{1}{24}((S^6)-3(S^4)+3(S^2)-1+3S^8-9S^6+9S^4-3S^2)+c_0$$ which simplifies to your second expression

Nyra
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