2

$\int^\pi_0\cos^6\theta\ d\theta$

So I split the trig value into:

$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$

Then I utilized the Pythagorean theorem for $cos^5\theta$

$\int^\pi_o(1-sin^5\theta)\ cos\theta$

I utilized u-substitution:

$u=sin\ \theta$

$du=cos\ \theta$

Thus:

$\int^{x=\pi}_{x=0}\ (1-u^5)\ d\theta$

I intergated $(\frac{1}{6}u^6)+(\frac{1}{6}u^6)$

$-(\frac{\pi^6}{6})+(0)$

$-(\frac{\pi^6}{6})$

Is my answer right?

Cetshwayo
  • 3,092

4 Answers4

1

Hint for question

$\int_0^{\pi}\cos^{2n} \theta d\theta=2\int_0^{\pi/2}\cos^{2n} \theta d\theta$

which is Wallis' integral: integral

Now for n=3 calculate the answer

Since $W_6=\frac{5\pi}{32} \Rightarrow \int_0^{\pi}\cos^{6} \theta d\theta=\frac{5\pi}{16}$

Angelo Mark
  • 5,954
0

Answer (using partial integration):

$$\int\cos^6(x)\space\space\text{d}x=$$ $$\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{6}\int\cos^4(x)\space\space\text{d}x=$$ $$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\cos^2(x)\space\space\text{d}x=$$ $$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\left(\frac{1}{2}\cos(2x)+\frac{1}{2}\right)\space\space\text{d}x=$$ $$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(2x)\space\space\text{d}x+\frac{5}{16}\int 1\space\space\text{d}x=$$

Substitute $u=2x$ and $\text{d}u=2\space\space\text{d}x$:

$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(u)\space\space\text{d}u+\frac{5}{16}\int 1\space\space\text{d}x=$$ $$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(u\right)}{32}+\frac{5x}{16}+\text{C}=$$ $$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(2x\right)}{32}+\frac{5x}{16}+\text{C}=$$ $$\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)+\text{C}$$

With boundaries:

$$\left[\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)\right]_{0}^{\pi}=$$ $$\frac{1}{192}\left[60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right]_{0}^{\pi}=$$ $$\frac{1}{192}\left(60\pi+45\sin(2\pi)+9\sin(4\pi)+\sin(6\pi)\right)=$$ $$\frac{1}{192}\left(60\pi+0+0+0\right)=\frac{60\pi}{192}=\frac{5\pi}{16}$$

Jan Eerland
  • 28,671
  • This question is from a trig substitution part of the text. One must be able to utilize that method to solve for this problem, no? – Cetshwayo Nov 09 '15 at 17:56
  • You can solve this problem multiple ways. If you definetely have to use trigonometry, try Andre Nicola's comment for using double-angle formula. – Ioannis Souldatos Nov 09 '15 at 17:59
0

$$\int^\pi_0\cos^6\theta\ d\theta=\int^\pi_0(\frac{1+\cos2\theta}{2})^3 d\theta$$ $$\frac{1}{8}\int^\pi_0(1+\cos2 \theta)^3d\theta=\frac{1}{8}\int^\pi_0(1+3\cos2\theta+3\cos^2 2\theta+\cos^32\theta)d\theta $$ $$\frac{1}{8}\int^\pi_0(1+3\cos2\theta+\frac{3}{2}(1+cos 4\theta)+\cos2\theta(1-\sin^22\theta)))d\theta$$ see, all terms can be integrated directly

E.H.E
  • 23,280
0

An easy way is to use Euler formula: $$\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$ Then use the binomial expansion: $$\cos^6\theta = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - k)\theta}e^{-ik\theta} = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - 2k)\theta}.$$ If $k = 3$, $\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \pi.$ Otherwise $$\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \frac{1}{6 - 2k}(e^{i(6 - 2k)\pi} - 1) = 0$$ in view of $6 - 2k$ is always an even number. Therefore, $$\int_0^\pi \cos^6\theta d\theta = \frac{1}{2^6} \binom{6}{3}\pi = \frac{5}{16}\pi.$$

Zhanxiong
  • 14,040