# Evaluate the indefinite integral integrate of cos^5(3x)dx

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...

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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

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

= (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 .