MHB LCC 206 {review 7r13} Integral substitution

[karush]

$\tiny\text{LCC 206 {review 7r13} Integral substitution}$
$$I=\int_{0}^{\pi/4}
\cos^5\left({2x}\right)\sin^2 \left({2x}\right)\,dx= \frac{4}{105} $$
$\text{u substitution } $
\begin{align}\displaystyle
u& = 2x&
du&= 2 \ d{x}&
2(\pi/4) &=\pi/2
\end{align}
$\text{then } $
$$I=2\int_{0}^{\pi/2}
\cos^5\left({u}\right)\sin^2 \left({u}\right)\,du
\implies \int_{0}^{\pi/2}
{\left(\cos^2 \left({u}\right)\right)}^{2}
\sin^2 \left({u}\right)\cos\left({u}\right) \ dx $$
$\text{Not sure if this is a good direction to go in?} $

$\tiny\text{ Surf the Nations math study group}$

 

[Prove It]

$\tiny\text{LCC 206 {review 7r13} Integral substitution}$
$$I=\int_{0}^{\pi/4}
\cos^5\left({2x}\right)\sin^2 \left({2x}\right)\,dx= \frac{4}{105} $$
$\text{u substitution } $
\begin{align}\displaystyle
u& = 2x&
du&= 2 \ d{x}&
2(\pi/4) &=\pi/2
\end{align}
$\text{then } $
$$I=2\int_{0}^{\pi/2}
\cos^5\left({u}\right)\sin^2 \left({u}\right)\,du
\implies \int_{0}^{\pi/2}
{\left(\cos^2 \left({u}\right)\right)}^{2}
\sin^2 \left({u}\right)\cos\left({u}\right) \ dx $$
$\text{Not sure if this is a good direction to go in?} $

$\tiny\text{ Surf the Nations math study group}$

Yes that's exactly the direction to go in, but you have made a mistake, it should be a factor of 1/2 out front, not 2. Now write $\displaystyle \begin{align*} \cos^2{(u)} = 1 - \sin^2{(u)} \end{align*}$ making the integral

$\displaystyle \begin{align*} \frac{1}{2}\int_0^{\frac{\pi}{2}}{\left[ \cos^2{(u)} \right] ^2\sin^2{(u)}\cos{(u)}\,\mathrm{d}u} &= \frac{1}{2}\int_0^{\frac{\pi}{2}}{ \left[ 1 - \sin^2{(u)} \right] ^2\sin^2{(u)} \cos{(u)}\,\mathrm{d}u } \\ &= \frac{1}{2}\int_0^{\frac{\pi}{2}}{ \left[ 1 - 2\sin^2{(u)} + \sin^4{(u)} \right] \sin^2{(u)} \cos{(u)}\,\mathrm{d}u } \\ &= \frac{1}{2} \int_0^{\frac{\pi}{2}}{ \left[ \sin^2{(u)} - 2\sin^4{(u)} + \sin^6{(u)} \right] \cos{(u)}\,\mathrm{d}u } \end{align*}$

Now you can make an appropriate substitution :)

 

[MarkFL]

Well, you would actually have:

$$I=\frac{1}{2}\int_0^{\frac{\pi}{2}} \cos^5(u)\sin^2(u)\,du$$

And then your observation that you can write:

$$I=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\cos^2(u)\right)^2\sin^2(u)\cos(u)\,du$$

Is a good one because now (using a Pythagorean identity) you can write the integral in the form:

$$I=\frac{1}{2}\int_0^{\frac{\pi}{2}} f(\sin(u))\,d(\sin(u))$$

 

[karush]

What would $f$ and $d$ be?

 

[MarkFL]

What would $f$ and $d$ be?

Your integrand and differential, respectively. :)

I fiddled around too long trying to make something work with the symmetry of the given integrand, and was beaten to the punch. :D

 

[karush]

$\displaystyle \begin{align*} \frac{1}{2}\int_0^{\frac{\pi}{2}}{\left[ \cos^2{(u)} \right] ^2\sin^2{(u)}\cos{(u)}\,\mathrm{d}u} &= \frac{1}{2}\int_0^{\frac{\pi}{2}}{ \left[ 1 - \sin^2{(u)} \right] ^2\sin^2{(u)} \cos{(u)}\,\mathrm{d}u } \\ &= \frac{1}{2}\int_0^{\frac{\pi}{2}}{ \left[ 1 - 2\sin^2{(u)} + \sin^4{(u)} \right] \sin^2{(u)} \cos{(u)}\,\mathrm{d}u } \\ &= \frac{1}{2} \int_0^{\frac{\pi}{2}}{ \left[ \sin^2{(u)} - 2\sin^4{(u)} + \sin^6{(u)} \right] \cos{(u)}\,\mathrm{d}u } \end{align*}$

$$\frac{1}{2}\left[
\frac{ \sin^3{(u)}}{3}-
\frac{ \sin^5{(u)}}{5}+
\frac{ \sin^7{(u)}}{7}\right]_0^{π/2}
=\frac{4}{105}$$