# `int_0^((1/6)pi) (2cos^3x- cos x)dx`

If we find hard to memorize formulas, we can deal with this integral in this way:

`int_0^(pi/6) (2cos^3x - cosx)dx = int_0^(pi/6) (cos^3x + cos^3x - cosx - cos x + cos x)dx`

Let's see what we have done: `2cos^3 x`  is written as cos^3x + cos^3x, added and subtracted then cos x.

`int_0^(pi/6) (2cos^3x - cosx)dx = 2int_0^(pi/6)(cos^3x- cosx)dx + int_0^(pi/6) cos xdx`

We can take out the factor cos x in `cos^3x- cosx.`

cos^3x - cosx = cosx(cos^2 x - 1)

The subtraction within brackets gives us the idea to put -`sin^2 x`  in stead of `cos^2 x - 1` .

`cos^3x- cosx = -cosx*sin^2 x`

`2int_0^(pi/6)(cos^3x- cosx)dx = 2int_0^(pi/6)(-cosx*sin^2 x)dx`

The substitution of `sin x`  with another variable works just perfect.

`sin x = u => (d(sinx))=du`

`2int_0^(pi/6)(-cosx*sin^2 x)dx = 2int_0^(1/2)u^2(-du)`

`2int_0^(pi/6)(-cosx*sin^2 x)dx = -2(1/2)^3/3 + 2*0^3/3`

`2int_0^(pi/6)(-cosx*sin^2 x)dx = -1/12`

We have to calculate one more integral `int_0^(pi/6) cos xdx` .

`int_0^(pi/6) cos xdx = sin(pi/6) - sin 0`

`int_0^(pi/6) cos xdx = 1/2`

`int_0^(pi/6) (2cos^3x - cosx)dx = -1/12 + 1/2`

`int_0^(pi/6) (2cos^3x - cosx)dx = (6-1)/12`

`int_0^(pi/6) (2cos^3x - cosx)dx = 5/12`

Answer: `int_0^(pi/6) (2cos^3x - cosx)dx = 5/12`

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We know that;

`cos 3x = 4cos^3 x -3 cosx`

`cos^3 x= (cos3x+3cosx)/4`

`int_0^((1/6)pi) (2cos^3 x-cosx)dx`

`= int_0^((1/6)pi) (2/4(cos3x+3cosx) -cosx)dx`

`= int_0^((1/6)pi)1/2 cos3x dx + (3/2-1)int_0^((1/6)pi) cosx dx`

`=1/2 xx 1/3 [sin3x]_0^(pi/6)+1/2[sinx]_0^(pi/6)`

`= 1/6(sin(pi/2)-sin0)+1/2(sin(pi/6)-sin0)`

`=1/6 xx 1 + 1/2 xx 1/2`

`=(2+3)/12`

`=5/12`