1. ## intergration

y= (x/ √ a^2 - x^2 ) - sin^-1 (x-a) (integrate)

∫ ln x /x^2 dx

∫ cos^6 1/2x dx

umm how do i do these ! pleases show me with working

thank you

2. ## Re: intergration

In the second expression, integrate by parts $\displaystyle \int v u' = vu - \int uv'$

Use $\displaystyle v = \ln(x) , u' = \frac{1}{x^2}$

3. ## Re: intergration

Hello, arsenal12345!

Here's the third one . . .

$\int \cos^6\!\tfrac{x}{2}\,dx$

$\cos^6\!\tfrac{x}{2} \;=\;\left(\cos^2\!\tfrac{x}{2}\right)^3 \;=\;\left(\tfrac{1 + \cos x}{2}\right)^3 \;=\;\tfrac{1}{8}\left(1 + 3\cos x + 3\cos^2\!x + \cos^3\!x\right)$

. . . . $=\;\tfrac{1}{8}\left(1 + 3\cos x + 3\left[\tfrac{1+\cos2x}{2}\right] + \cos^2\!x\!\cdot\cos x\right)$

. . . . $=\;\tfrac{1}{8}\left(1 + 3\cos x + \tfrac{3}{2} + \tfrac{3}{2}\cos2x + [1-\sin^2\!x]\cos x\right)$

. . . . $=\;\tfrac{1}{8}\left(\tfrac{5}{2} + 3\cos x + \tfrac{3}{2}\cos2x + \cos x - \sin^2\!x\cos x\right)$

. . . . $=\;\tfrac{1}{8}\left(\tfrac{5}{2} + 4\cos x +\tfrac{3}{2}\cos2x - \sin^2\!x\cos x\right)$

You should be able to handle those four integrals.