0
$\begingroup$

How does integral of $(\sin x)^2 (\cos x)^3 = {1\over5}(\sin x)^5 - {1\over3}(\sin x)^3$ manage to turn into ${1\over30}(\sin x)^3 (3\cos(2x)+7)$?

  • 0
    There is a sign error. Use the fact that $\cos 2x=1-2\sin^2 x$.2011-09-15
  • 2
    Title mentions integral, body doesn't. Please edit one or the other so they match.2011-09-15
  • 1
    Are the $\sin x$'s in the denominators or numerators? Is it $\frac{1}{5 (\sin x)^5}-\frac{1}{3 (\sin x)^3}$ or $\frac{(\sin x)^5}{5 }-\frac{(\sin x)^3}{3 }$?2011-09-15
  • 0
    I've made some guesses as to what was meant and put in a little TeX. I think Andre's comment is exactly right (and wonder whether Andre would consider making it an answer).2011-09-15
  • 0
    @Gerry Myerson: OK, done. In gratitude, I will try to extend a similar invitation to you.2011-09-15
  • 0
    I don't know why but these are questions for online homework and it insists on you manipulating the answer every time to get the exact answer they have instead of allowing for any answer that is it's equivalent.2011-09-15
  • 0
    Concerning the integral I do not understand your revised question. Which is the the integral you want to evaluate? It can be proved that $$\begin{eqnarray*} \int (\sin x)^{2}(\cos x)^{3}dx &=&-\frac{1}{5}\cos ^{4}x\sin x+\frac{1}{15} \cos ^{2}x\sin x+\frac{2}{15}\sin x \\ &\neq &\frac{1}{5}\sin ^{5}x-\frac{1}{3}\sin ^{3}x, \end{eqnarray*}$$ while $$\begin{eqnarray*} \int \left( -\cos ^{3}x+\cos ^{5}x\right) dx &=&\frac{1}{5}\sin ^{5}x-\frac{1 }{3}\sin ^{3}x \\ &=&-\frac{1}{30}(\sin x)^{3}\left( 3\cos 2x+7\right) . \end{eqnarray*}.$$2011-09-16

1 Answers 1

1

Let $$F(x)={1\over5}(\sin x)^5 - {1\over3}(\sin x)^3\qquad\text{(Equation 1)}$$ The right-hand side seems like an attractive enough expression. But if overtaken by the urge to manipulate, we might note the common factor $(\sin x)^3$. We may also want to bring the expression to the common denominator $15$. We arrive at the equation $$F(x)={1\over15}(\sin x)^3 (3(\sin x)^2-5)\qquad\text{(Equation 2)}$$ Now, if the urge to tinker is not yet spent, we may want to use the double-angle trigonometric identity $$\cos(2x)=(\cos x)^2-(\sin x)^2=1-2(\sin x)^2.$$ Rewrite the above identity as $$(\sin x)^2=\frac{1-\cos(2x)}{2},$$ and note that $$3(\sin x)^2-5=\frac{3(1-\cos(2x))}{2} -5=-\frac{3\cos(2x) +7}{2}.$$ Now substitute in Equation $2$. We obtain $$F(x)=-{1\over30}(\sin x)^3(3\cos(2x) +7).$$

Apart from a minus sign, this is the expression given in the question.

Comment: It is hard to imagine a reason for preferring the expression we arrived at over the original expression.

  • 0
    I suppose it has the advantage that you can immediately tell that it is zero if and only if either $\sin x=0$ or $\cos2x=-7/3$.2011-09-15