Trig Integral Example - # 2

[Jason76]

\(\displaystyle \int \cos^{2} x \sin x dx\)

\(\displaystyle u = \cos x\)

\(\displaystyle du = \sin x dx\)

\(\displaystyle -du = \sin x dx\)

\(\displaystyle -\int u^{2} du\)

\(\displaystyle \rightarrow -\dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow -\dfrac{\cos^{3} x}{3} + C\) This is right according to the book.

How does this compare to, regarding integration strategy (which I believe is wrong):

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x dx\)

\(\displaystyle \dfrac{1}{\sec x}du = \tan x dx\)

\(\displaystyle \dfrac{1}{\sec x} \int \sec^{2} x du\)

Using the formula \(\displaystyle \dfrac{u^{n + 1}}{n + 1}\) which we would only use in the case of trig function raised to some power.

\(\displaystyle \rightarrow (\dfrac{1}{\sec x})\dfrac{u}{3} + C\)

\(\displaystyle \rightarrow (\dfrac{1}{\sec x})\dfrac{\sec^{3} x}{3} + C\)

 

[Jason76]

Where does the \(\displaystyle \dfrac{1}{2}\) come from?

 

[Jason76]

Starting integral:

\(\displaystyle \int \sec^{2} x \tan x dx\)

\(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x\)

So how to manipulate the du so that it has \(\displaystyle \tan x dx\) on the right side (of the \(\displaystyle du = dx \) equation, as it is in the starting integral.

You said this has something to do with \(\displaystyle \dfrac{1}{2}\)

 

[Deleted member 4993]

Starting integral:



\(\displaystyle u = \sec x\)

\(\displaystyle du = \sec x \tan x\)dx

So how to manipulate the du so that it has \(\displaystyle \tan x dx\) on the right side (of the \(\displaystyle du = dx \) equation, as it is in the starting integral.

You said this has something to do with \(\displaystyle \dfrac{1}{2}\)

You do realize that you can also do:

u = tan(x)

du = sec²(x) dx

Then

\(\displaystyle \displaystyle{\int tan(x) * sec^2(x) \ \ dx \ = \ \int u \ \ du \ = \ \frac{1}{2}u^2 + C \ = \ \frac{1}{2}tan^2(x) + C}\)

 

[Jason76]

Ok, since I see du = dx manipulation won't work, then when \(\displaystyle u = \sec x\). then trig substitution is the only way to go. But what about when you make \(\displaystyle u = \tan x\)