Odd powers of trig integrands such as in $\int \sin^7 \theta\cos^5\theta\, d\theta$

Question

I am doing as many trig integrals exercices as I can to develop the skills and thought patterns. I have a rather simple question about the result I get when computing integrals such as $\int \sin^7 \theta\cos^5\theta\, d\theta$, I use the basic substitution technique

$$\int \sin^7 \theta\cos^5\theta\, d\theta\,=\,\int u^7\theta\left(1-u^2\theta\right)^2\, du\,:\,u=\sin\theta \quad du=\cos\theta\, d\theta$$ wich gives me $$\frac 18 \sin^8\theta - \frac 15 \sin^{10}\theta + \frac 1{12} \sin^{12}\theta +C$$ I check my answers with $Mathematica$ which gives, $$\frac {-5\cos{2\theta}}{1024}+\frac {5\cos{4\theta}}{8192}+\frac {5\cos{6\theta}}{6144} - \frac {\cos{8\theta}}{4096} - \frac {\cos{10\theta}}{10240}+\frac {\cos{12\theta}}{24 576}$$ this is far from the result I got by substitution so I thought I was doing something wrong and I tried to achieve the same result but it is a bit painful, so I tried a definite integral as a quicker way to convince myself. Sure enough, I get the same answer as $Mathematica$ when I compute definite integrals. For example, $$\int_0^\frac \pi 2 \sin^7 \theta\cos^5\theta\, d\theta\,=\,\frac1{120}$$ $$\frac 18 \sin^8\frac \pi 2- \frac 15 \sin^{10}\frac \pi2+ \frac 1{12} \sin^{12}\frac \pi2\,=\,\frac 1{120}$$ Does this have to do with the way $Mathematica$ does computations or am I doing something wrong? Also, is it useful to get it down to the same form as does $Mathematica$ or is the simple u-substitution result always reliable and correct?

Answer 1

Actually, both of your solutions are correct. They are just of a different form.

Using the following identities: $$\cos(2\theta)\equiv \cos^2 \theta-\sin^2 \theta \tag{1}$$ $$\sin(2\theta)\equiv 2\sin \theta \cos \theta \tag{2}$$ $$\sin^2 \theta+\cos^2 \theta\equiv 1 \tag{3}$$ We can obtain the result you've obtained from Mathematica: $$\frac 18 \sin^8\theta - \frac 15 \sin^{10}\theta + \frac 1{12} \sin^{12}\theta +C$$ To do so, replace each of the $\cos$ terms on the result you've obtained from Mathematica with powers of $\sin \theta$. $$\frac {-5\cos(2\theta)}{1024}+\frac {5\cos(4\theta)}{8192}+\frac {5\cos(6\theta)}{6144} - \frac {\cos(8\theta)}{4096} - \frac {\cos(10\theta)}{10240}+\frac {\cos(12\theta)}{24 576}$$

For the $\cos(4\theta), \cos(6\theta),\cdots ,\cos(12\theta)$ terms apply all the 3 above identities:

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This is an example for $\cos(4\theta)$: $$\cos(4\theta)=\cos^2(2\theta)-\sin^2(2\theta)=(\cos^2 \theta-\sin^2\theta)^2-(2\sin \theta\cos\theta)^2$$ $$\therefore \cos(4\theta)=\sin^4\theta+\cos^4\theta-6\sin^2 \theta\cos^2 \theta $$ $$\therefore \cos(4\theta)=\sin^4\theta+(1-\sin^2\theta)^2-6\sin^2\theta\cdot (1-\sin^2\theta)$$ $$\therefore \cos(4\theta)=8\sin^4\theta-8\sin^2 \theta+1$$ Now apply this process to the other $\cos$ terms.

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If you would not like to do this method by hand (since it is extremely tedious), you can check this using Wolfram|Alpha. Do not be alarmed that it suggests that no solutions exist on this specific input. That is because the arbitrary constant of integration $C$ was not considered (It was assumed that $C=0$). If you let $C=-\frac{462}{122880}$, you will see that they are identical and apply for all $\theta \in \mathbb{R}$.

Answer 2

$$ \begin{aligned} \int \sin ^7 \theta \cos ^5 \theta d \theta = & \int \frac{\sin^5 (2 \theta)}{32} \sin ^2 \theta d \theta \\ = & \frac{1}{32} \int \sin ^5(2 \theta)\left(\frac{1-\cos (2 \theta)}{2}\right) d \theta \\ = & \frac{1}{64}\left[\int \sin ^5(2 \theta) d \theta-\int \sin ^5(2 \theta) \cos (2 \theta) d \theta\right] \\ = & \frac{1}{64}\left[-\frac{1}{2} \int\left(1-\cos ^2(2 \theta)\right)^2 d(\cos (2 \theta))-\frac{1}{2} \int \sin ^5(2 \theta) d(\sin (2 \theta))\right] \\ = & -\frac{1}{128}\left[\cos (2 \theta)-\frac{2 \cos ^3(2 \theta)}{3}+\frac{\cos ^5(2 \theta)}{5}-\frac{\sin ^6(2 \theta)}{6}\right]+C \end{aligned} $$