Question

how to integrate sin^(8) (x)? i keep on getting the cos^4 x and my solution is so long and i don't know how to end it. i know that i can reduce this sin^8 x into half angle identity but i don't know how to continue.. please help

Answer 1

You're going to be using the reduction formula:\[\int \sin^n(x)dx = -\frac{\cos(x)\sin^{n-1}(x)}{n}+\frac{n-1}{n}\int \sin^{n-2}(x)dx\]

Answer 2

where \(n=8\)

Answer 3

You will use this formula a couple times till you can get your integral: \(\int \sin^{n-2}(x)dx\) to the second power.

Answer 4

I say listen to Jhanny. = w=

Answer 5

Thanks for your help but we are not allowed to use the reduction formula. Thanks a lot! :D

Answer 6

i did get the answer by using the half angle identity.

Answer 7

Oh you're not?? tHat's a bummer!!

Answer 8

I can see how the half angle identity would also work here.

Answer 9

\[\begin{align*}

\sin^8x&=\left(\frac{1-\cos2x}{2}\right)^4\\\\

&=\frac{1}{16}\left[1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x\right]\\\\

&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)\right.\\
&\quad\quad\quad\quad\left.+\frac{1}{4}(1+\cos4x)^2\right]\\\\

&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)\right.\\
&\quad\quad\quad\quad\left.+\frac{1}{4}(1+2\cos4x+\cos^24x)\right]\\\\

&=\frac{1}{16}\left[1-4\cos2x+3(1+\cos4x)-4\cos2x(1-\sin^22x)\right.\\
&\quad\quad\quad\quad\left.+\frac{1}{4}(1+2\cos4x+\frac{1}{2}(1+\cos8x))\right]\\\\

&=\frac{1}{16}\left[\frac{35}{8}+\frac{7}{2}\cos4x-4\cos2x(2-\sin^22x)+\frac{1}{8}\cos8x\right]
\end{align*}\]
Each term can be integrated easily.