"Evaluate the following. ∫cos5(3z)dzsin2(3z)" was answered by students like you

Lennie Carroll

Answered question

2020-10-25

Evaluate the following.

\int \frac{\cos^{5} (3 z) d z}{\sin^{2} (3 z)}

Mitchel Aguirre

Skilled2020-10-26Added 94 answers

We have to find the integral of:

\int \frac{\cos^{5} (3 x)}{\sin^{2} (3 x)} d x

Substitute

u = 3 x \Rightarrow \frac{d u}{d x} = 3 \Rightarrow \frac{d u}{3}

, we obtain

\int \frac{\cos^{5} (3 x)}{\sin^{2} (3 x)} d x = \frac{1}{3} \int \frac{\cos^{5} (3 u)}{\sin^{2} (3 u)} d u

Now solving:

\int \frac{\cos^{5} (3 u)}{\sin^{2} (3 u)} d u = \int \frac{\cos^{5} (U)}{\sin^{2} (U)} d U

Preparing for substitution, we use:

\cos^{2} U = 1 - \sin^{2} U

, we obtain

\int \frac{\cos^{5} (U)}{\sin^{2} (U)} d U

= \int \cos (U) \frac{{(\sin^{2} (U) - 1)}^{2}}{\sin^{2} (U)} d U

Substitute

v = \sin (U) \Rightarrow \frac{d v}{d U} = \cos (u) \Rightarrow d u = \frac{1}{\cos (u)} d v

, we obtain:

= \int \frac{{(v^{2} - 1)}^{2}}{v^{2}} d v

= \int (v^{2} + \frac{1}{v^{2}} - 2) d v

= \int (v^{2}) d v + \int (\frac{1}{v^{2}}) d v - 2 \int d v

= \frac{v^{3}}{3} - \frac{1}{v} + v

Now, we can undo the substitution

v = \sin (u)

, we obtain:

= \frac{v^{3}}{3} - \frac{1}{v} - 2 v

= \frac{\sin^{3} (u)}{3} - \frac{1}{\sin (u)} - 2 \sin (u)

Plug in solved integrals,

This is helpful 8 Flag

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?