1. ## Trivial Integral

Given that $\displaystyle \int^\frac{\pi}{4}_0 \sin^2x\cos^6x\;dx=k$, find the value of $\displaystyle \int^\frac{\pi}{4}_0 \sin^4x\cos^6x\;dx$ in terms of $\displaystyle k$.

Thanks

2. Originally Posted by polymerase
Given that $\displaystyle \int^\frac{\pi}{4}_0 \sin^2x\cos^6x\;dx=k$, find the value of $\displaystyle \int^\frac{\pi}{4}_0 \sin^4x\cos^6x\;dx$ in terms of $\displaystyle k$.

Thanks
using $\displaystyle \sin^2(x)=1-\cos^2(x)$ we get

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx-\int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx$

lets focus on the 2nd integral for a second

$\displaystyle \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx$ let's use integration by parts with

$\displaystyle u=\cos^7(x) \to du=-7\sin(x)\cos^6(x)$ and

$\displaystyle dv=\sin^2(x)\cos(x)dx \to v=\frac{1}{3}\sin^3(x)$

$\displaystyle \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx=\frac{1}{3}\sin^3(x)\cos^7( x)\bigg|_{0}^{\pi/4} +\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=$

$\displaystyle \frac{1}{3} \left( \frac{1}{\sqrt{2}}\right)^3\left( \frac{1}{\sqrt{2}}\right)^7+\frac{7}{3}\int_{0}^{\ pi/4}\sin^4(x)\cos^6(x)dx=\frac{1}{96}+\frac{7}{3}\in t_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$

Now subbing this back into the original we get

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \left[ \frac{1}{96}+\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx\right]$

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}-\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$

Now we add $\displaystyle \frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$ to both sides of the equation to get

$\displaystyle \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}$

We know that the integral on the right hand side is k so we get

$\displaystyle \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=k - \frac{1}{96}$

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\frac{3}{10} \left( k - \frac{1}{96} \right) =\frac{3}{10}k-\frac{1}{320}$

Yippy !!

3. Originally Posted by TheEmptySet
using $\displaystyle \sin^2(x)=1-\cos^2(x)$ we get

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx-\int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx$

lets focus on the 2nd integral for a second

$\displaystyle \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx$ let's use integration by parts with

$\displaystyle u=\cos^7(x) \to du=-7\sin(x)\cos^6(x)$ and

$\displaystyle dv=\sin^2(x)\cos(x)dx \to v=\frac{1}{3}\sin^3(x)$

$\displaystyle \int_{0}^{\pi/4}\sin^2(x)\cos^8(x)dx=\frac{1}{3}\sin^3(x)\cos^7( x)\bigg|_{0}^{\pi/4} +\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=$

$\displaystyle \frac{1}{3} \left( \frac{1}{\sqrt{2}}\right)^3\left( \frac{1}{\sqrt{2}}\right)^7+\frac{7}{3}\int_{0}^{\ pi/4}\sin^4(x)\cos^6(x)dx=\frac{1}{96}+\frac{7}{3}\in t_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$

Now subbing this back into the original we get

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \left[ \frac{1}{96}+\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx\right]$

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}-\frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$

Now we add $\displaystyle \frac{7}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx$ to both sides of the equation to get

$\displaystyle \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\int_{0}^{\pi/4}\sin^2(x)\cos^6(x)dx- \frac{1}{96}$

We know that the integral on the right hand side is k so we get

$\displaystyle \frac{10}{3}\int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=k - \frac{1}{96}$

$\displaystyle \int_{0}^{\pi/4}\sin^4(x)\cos^6(x)dx=\frac{3}{10} \left( k - \frac{1}{96} \right) =\frac{3}{10}k-\frac{1}{320}$

Yippy !!
but the answers says...$\displaystyle \frac{3k}{10}-\frac{1}{320}$

4. Originally Posted by polymerase
but the answers says...$\displaystyle \frac{3k}{10}-\frac{1}{320}$
Um...that's exactly the answer that TES got...

(3/10)*k = (3k)/10

5. I had a typo but I fixed it. He must have saw it then

6. Originally Posted by TheEmptySet
I had a typo but I fixed it. He must have saw it then
Oh, ok.

why're you trying to make polymerase look bad with fractions ?!

7. Originally Posted by Jhevon
Oh, ok.

why're you trying to make polymerase look bad with fractions ?!
I guess I am just EVIL at heart....

He must have looked right after I posted becuase I noticed right after and edited it. It is really ironic. I did all of the problem but I get to the end and need to simplify $\displaystyle \left( \frac{1}{\sqrt{2}}\right)^{10}$ and I guess I forgot how to use exponents .

Brain glitch I guess.