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

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There is a sign error. Use the fact that $\cos 2x=1-2\sin^2 x$. –  André Nicolas Sep 15 '11 at 2:47
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Title mentions integral, body doesn't. Please edit one or the other so they match. –  Gerry Myerson Sep 15 '11 at 4:13
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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 $\frac{(\sin x)^5}{5 }-\frac{(\sin x)^3}{3 }$? –  Ross Millikan Sep 15 '11 at 4:38
    
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). –  Gerry Myerson Sep 15 '11 at 6:36
    
@Gerry Myerson: OK, done. In gratitude, I will try to extend a similar invitation to you. –  André Nicolas Sep 15 '11 at 7:48

1 Answer 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.

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I suppose it has the advantage that you can immediately tell that it is zero if and only if either $\sin x=0$ or $\cos2x=-7/3$. –  Gerry Myerson Sep 15 '11 at 9:17

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