"Integral Calculus: Trigonometry and Inverse Trigonometry. Evuate the..." solved and explained

compagnia04

Answered question

2021-12-14

Integral Calculus: Trigonometry and Inverse Trigonometry. Evuate the following.

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

kaluitagf

Beginner2021-12-15Added 38 answers

Step 1
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

Step 2 Substitute

v = \sin (U) \Rightarrow \frac{d v}{d U} \Rightarrow \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}}) - 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,

Annie Levasseur

Beginner2021-12-16Added 30 answers

The site is very convenient and everything is accessible and understandable, thanks to this I was able to reach a new level in mathematics

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