1. plz help! Integration problems

Alright. I already put a lot of thought into some of these but I cannot get the right answer out or am stuck in the middle.

1) The integral, from 0 to pi/4, of ((sinx)^4)((cosx)^2)dx

My thought process has gotten to (1/16)integral(1-cos4x)dx -integral((1/8)((cosx)^2)((sin2x)^2)dx

What do i do from here? Or is there a different way to go at it?

2) Integral, from pi/4 to pi/2 of (cotx)^2 for this one, I got all the way to solving the integral, which i Think is -.5(cotx)^2 -ln(abs value of)sinx ...

I am stuck here, since pi/2 cant go into cotangent... what do I do?

3) The integral of 1 to radical 3 of arctan(1/x)dx. I solved this several times, and I got the integral to come out as xarctan(1/x) +.5ln(x^2 +1)
However, this is wrong, because i know that the answer SHOULD be -.468. How do i get -.468?

2. 1. I'll have to play with this one ... even powers of sine and cosine are a pain-in-the-...

2. $\cot^2{x} = \csc^2{x} - 1$

3. how is it negative ?

from 1 to sqrt(3), the function y = arctan(1/x) is greater than 0.

I get positive .468

3. Hello, 3deltat!

$3)\;\;\int^{\sqrt{3}}_1 \arctan\left(\frac{1}{x}\right)\,dx$

I got the integral to come out as: . $x\arctan\left(\frac{1}{x}\right) + \frac{1}{2}\ln(x^2 +1)$ . . . . Right!

However, this is wrong, because i know that the answer SHOULD be -0.468 . ??
How can the answer be negative? . . . The graph is above the x-axis.
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Ah, I see skeeter already beat me to it!

We have: . $x\arctan\left(\frac{1}{x}\right) + \frac{1}{2}\ln(x^2+1)\:\bigg]^{\sqrt{3}}_1$

. . $\bigg[\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\right) + \frac{1}{2}\ln(4)\bigg] - \bigg[1\!\cdot\!\arctan(1) + \frac{1}{2}\ln(2)\bigg]$

. . $= \;\bigg[\sqrt{3}\left(\frac{\pi}{6}\right) + \ln\left(4^{\frac{1}{2}}\right)\bigg] - \bigg[\frac{\pi}{4} + \frac{1}{2}\ln(2)\bigg]$

. . $= \;\frac{\pi\sqrt{3}}{6} + \ln(2) - \frac{\pi}{4} - \frac{1}{2}\ln(2)$

. . $= \;\left(\frac{2\sqrt{3}-3}{12}\right)\pi + \frac{1}{2}\ln(2)$

. . $= \;\boxed{0.468}075109$

4. $\int \sin^4{x}\cos^2{x}$

Here's my way:

$\frac{1}{8} \int (1-\cos{2x})^2(1+\cos{2x})$

$\frac{1}{8} \int (\underbrace{1-\cos^2{2x}}_{\sin^2{2x}})(1-\cos{2x})$

$\frac{1}{8} \int (\sin^2{2x} - \sin^2{2x}\cos{2x})$

And we're done...

Another double angle formula for the first integral and sub $u = \sin{2x}$ for second integral.

5. Originally Posted by 3deltat
Alright. I already put a lot of thought into some of these but I cannot get the right answer out or am stuck in the middle.

1) The integral, from 0 to pi/4, of ((sinx)^4)((cosx)^2)dx

My thought process has gotten to (1/16)integral(1-cos4x)dx -integral((1/8)((cosx)^2)((sin2x)^2)dx

What do i do from here? Or is there a different way to go at it?
When both are even, you need to apply these identities: $\sin^2 u=\frac{1-\cos (2u)}{2}$ and $\cos^2u=\frac{1+\cos(2u)}{2}$:

$\int_0^{\frac{\pi}{4}}\left[\sin^2x\right]^2\cos^2x\,dx=\int_0^{\frac{\pi}{4}}\left[\tfrac{1}{2}(1-\cos (2x))\right]^2\left[\tfrac{1}{2}(1+\cos(2x))\right]\,dx$ $=\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[1-\cos (2x)-\cos^2(2x)+\cos^3(2x)\right]\,dx$

Now split up the integral:

$\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[1-\cos (2x)-\cos^2(2x)+\cos^3(2x)\right]\,dx$ $=\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[1-\cos (2x)-\cos^2(2x)\right]\,dx + \tfrac{1}{8}\int_0^{\frac{\pi}{4}}\cos^3(2x)\,dx$

Let's focus on this integral:

$\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[1-\cos (2x)-\cos^2(2x)\right]\,dx$

This is the same as saying $\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[\tfrac{1}{2}-\cos (2x)-\tfrac{1}{2}\cos(4x)\right]\,dx$

Evaluating, we get $\tfrac{1}{8}\left.\left[\tfrac{1}{2}x-\frac{1}{2}\sin(2x)-\tfrac{1}{8}\sin(4x)\right]\right|_0^{\frac{\pi}{4}}=\tfrac{1}{8}\left[\tfrac{1}{8}\pi-\tfrac{1}{2}\right]=\frac{\pi-4}{64}$

Now let's evaluate $\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\cos^3(2x)\,dx$

Break off a factor of $\cos(2x)$ and apply the identity $1-\sin^2 u = \cos^2 u$

Thus, the integral becomes $\tfrac{1}{8}\int_0^{\frac{\pi}{4}}\left[1-\sin^2(2x)\right]\cos(2x)\,dx$

Now let $z=\sin(2x)\implies \,dz=2\cos(2x)\,dx$

We can change the limits of integration as well.

The integral can now be written as $\tfrac{1}{16}\int_0^1\left[1-z^2\right]\,dz=\tfrac{1}{16}\left.\left[z-\tfrac{1}{3}z^3\right]\right|_0^1=\frac{2}{48}$

Finally, our total solution is $\frac{\pi}{64}-\frac{1}{48}=\color{red}\boxed{\frac{3\pi-4}{192}}$

Does this make sense?

--Chris

6. Originally Posted by Chop Suey
$\int \sin^4{x}\cos^2{x}$

Here's my way:

$\frac{1}{8} \int (1-\cos{2x})^2(1+\cos{2x})$

$\frac{1}{8} \int (\underbrace{1-\cos^2{2x}}_{\sin^2{2x}})(1-\cos{2x})$

$\frac{1}{8} \int (\sin^2{2x} - \sin^2{2x}\cos{2x})$

And we're done...

Another double angle formula for the first integral and sub $u = \sin{2x}$ for second integral.
mmmmmk...

This is a "bit" easier

--Chris

7. Yep this makes more sense! Thank you for your help. =)