How does integral of $(\sin x)^2 (\cos x)^3 = {1\over5}(\sin x)^5 - {1\over3}(\sin x)^3$ manage to turn into ${1\over30}(\sin x)^3 (3\cos(2x)+7)$?
Definite Integral with trigonometric identity answer
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0Concerning 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
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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0I 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