Indefinite Trig Integrals: int (sin^4x)(cos^2x) dx and

[maeveoneill]

I am doing a chapter on trigonometic integrals. I have just come across a couple questions that I don't know how to rearrange at the beginning to solve for the integral..

1. integral of (sin^4x)(cos^2x) dx

Because the powers are both even, the book suggests using half angle formulas.


2. integral of (sin^3x)(sqrt cosx) dx

If someone could just help me with the first step of rearranging in order to solve that would be great!

 

[skeeter]

\(\displaystyle \L \sin^4{x}\cos^2{x} =\)

\(\displaystyle \L (\sin^2{x})^2\cos^2{x} =\)

\(\displaystyle \L \left(\frac{1-\cos{(2x)}}{2}\right)^2 \left(\frac{1 + \cos{(2x)}}{2}\right) =\)

\(\displaystyle \L \left(\frac{1 - cos^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right)\)

\(\displaystyle \L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =\)

\(\displaystyle \L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =\)

\(\displaystyle \L \frac{1}{8} \left[\frac{1 - \cos{(4x)}}{2} - \sin^2{(2x)}\cos{(2x)}\right] =\)

you now should be able to integrate the last expression.



for the 2nd one ...

\(\displaystyle \L \sin^3{x} \cdot \sqrt{\cos{x}} =\)

\(\displaystyle \L \sin{x}(1 - cos^2{x})\sqrt{\cos{x}} =\)

\(\displaystyle \L \left(\cos^{\frac{5}{2}}{x} - \cos^{\frac{1}{2}}{x}\right)(-\sin{x})\)

finish up.

 

[maeveoneill]

how did you get from this step to the next?

from here \(\displaystyle \L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =\)

to here \(\displaystyle \L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =\)


.. and thank you.. i was able to figure out the 2nd one!

 

[skeeter]

how did you get from this step to the next?

from here \(\displaystyle \L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =\)

to here \(\displaystyle \L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =\)

no big mystery ... I factored out the two constants in the denominator and multiplied (distributed) the numerators ...
sin²(2x)[1 - cos(2x)] = sin²(2x) - sin²(2x)cos(2x)


.. and thank you.. i was able to figure out the 2nd one!

 

[maeveoneill]

then you would use substitution, u = sin2x and du= 1/2cos2x dx
would your du not cancel out the cos 2x in the equation and then elave you with u^2 - u^2?? = 0?

 

[skeeter]

\(\displaystyle \L \frac{1}{8}\int \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] dx\)

\(\displaystyle \L \frac{1}{8}\int \sin^2{(2x)} dx - \frac{1}{8}\int \sin^2{(2x)}\cos{(2x)}dx\)

\(\displaystyle \L \frac{1}{8} \int \frac{1 - \cos{(4x)}}{2} dx - \frac{1}{16}\int \sin^2{(2x)} \cdot 2\cos{(2x)}dx\)

\(\displaystyle \L \frac{1}{16} \int 1 - \cos{(4x)} dx - \frac{1}{16}\int u^2 du\)

\(\displaystyle \L \frac{1}{16}\left[x - \frac{\sin{(4x)}}{4} - \frac{\sin^3{(2x)}}{3}\right] + C\)