Evaluate the indefinite integral integrate of cos^5(3x)dx
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`int cos^5(3x) dx`
To start, let's use substitution method.
So let,
`u =3x`
Then, differentiate u.
`du = 3dx`
`(du)/3 = dx`
And replace the x variable in the integral with u.
`int cos^5 u (du)/3 = 1/3 int cos ^5 u du`
Next, express the cosine function with a power of 1. To do so, apply the Pythagorean identity `sin^2 theta + cos^2 theta =1` .
`= 1/3 int cos u * cos^2 u*cos^2 u du`
`= 1/3 int cos u(1-sin^2u)(1-sin^2u)du`
`=1/3 int cos u(1 -2sin^2u+sin^4u)du`
`=1/3[ int cos udu -2 int sin^2u cos udu + int sin^4u cos u du]`
And then, let's evaluate each integral separately.
>>> For the first integral, apply the formula `int cos u = sin u + C` .
`int cos u du = sin u + C`
>>> For the second integral `int sin^2 u cos u du` , use substitution again. So let,
`y = sin u`
Then,
`dy = cos u du`
So replacing the u variable with y yields:
`int y^2 dy= y^3/3 + C`
To return to variable u, substitute back y =sin u. Hence,
`int sin^2 u cos u du = (sin^3u)/3 + C`
>>> For the third integral `int sin^4 u cos u du` , use substitution method too. So let,
`z=sin u`
Then, differentiate z.
`dz=cos u du`
And replace the u variable with z.
`int z^4 dz = z^5/5 + C`
Substitute back z=sin u. Hence,
`int sin^4 u cos u du = (sin^5u)/5 + C`
Then we have:
`1/3[intcos u du -2int sin^2u du + int sin^4 cos u du]`
`= 1/3 [ sin u + C - 2((sin^3u)/3 + C)+(sin^5 u) /5 + C ]`
Since C is any constant, we may represent the three C's as C only.
`= 1/3[sin u-(2sin^3u)/3 + (sin^5u)/5 + C]`
`= (sin u)/3 - (2sin^3u)/9 + (sin^5 u)/ 15 + C`
To return to the original variable, substitute u=3x.
`= (sin(3x))/3 - (2sin^3(3x))/9+(sin^5(3x))/15 + C`
Hence, `int cos^5(3x) dx = (sin (3x))/3 - (2sin^3(3x))/9 + (sin^5 (3x))/15 + C` .
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