How can one integral have two answers? 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?


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
