Author Topic: Nagisa Maths Thread (Read 21320 times) Tweet Share

Nagisa · « **Reply #30 on:** December 26, 2012, 01:40:35 am »

Quote from: Hancock on December 26, 2012, 01:26:50 am

$\int e^{2x}\sin{3x} \ dx$

$Im(\int e^{2x}e^{3ix} \ dx)$

$Im(\int e^{(2+3i)x}\ dx)$

$Im(\frac{1}{2+3i}e^{(2+3i)x})$

$Im(\frac{2-3i}{13}e^{2x}e^{3ix})$

$Im(\frac{2-3i}{13}e^{2x}(cos(3x) + isin(3x))$

$\frac{e^{2x}}{13}Im((2-3i)(cos(3x) + isin(3x))$

$\frac{e^{2x}}{13}(2sin(3x)-3cos(3x))$

im learning integration by parts atm so i was wanting a solution by that method...

anyways this is what ive done and i think im stuck but ye #yolo

$u = e^{2x} \ \ \ dv = \sin 3x \ dx$

$du = 2e^{2x} \ dx \ \ \ v = -\cos 3x . \frac{1}{3}$

so $e^{2x} \sin 3x \ dx = e^{2x}\left ( \frac{-\cos3x}{3}\right) - \int 2e^{2x} \left ( \frac{-\cos3x}{3}\right) \ dx$

$\left ( \frac{-e^{2x}\cos3x}{3}\right) + \left ( \frac{e^{2x}\sin3x}{9}\right)$

so then i did this $\frac{e^{2x}(3\sin3x - 9\cos3x)}{27}$

help plss

Hancock · « **Reply #31 on:** December 26, 2012, 01:59:18 am »

$\int e^{2x}\sin{3x} \ dx$

$u = e^{2x}$ $\frac{du}{dx} = 2e^{2x}$
$\frac{dv}{dx} = sin(3x)$ $v = \frac{-1}{3}cos(3x)$

$\int e^{2x}\sin{3x} \ dx = \frac{-1}{3}e^{2x}cos(3x) - \int \frac{-2}{3}e^{2x}cos(3x) dx$

$\int e^{2x}\sin{3x} \ dx = \frac{-1}{3}e^{2x}cos(3x) + \frac{2}{3}\int e^{2x}cos(3x) dx$

Need to do integration by parts again on the integral to the right.

$u = e^{2x}$ $\frac{du}{dx} = 2e^{2x}$
$\frac{dv}{dx} = cos(3x)$ $v = \frac{1}{3}sin(3x)$

$\int e^{2x}cos(3x) dx = \frac{1}{3}e^{2x}sin(3x) - \int \frac{2}{3}e^{2x}sin(3x) dx$

$\int e^{2x}cos(3x) dx = \frac{1}{3}e^{2x}sin(3x) - \frac{2}{3}\int e^{2x}sin(3x) dx$

Putting this back into the top equation:

$\int e^{2x}\sin{3x} \ dx = \frac{-1}{3}e^{2x}cos(3x) + \frac{2}{3}(\frac{1}{3}e^{2x}sin(3x) - \frac{2}{3}\int e^{2x}sin(3x) dx$

$\int e^{2x}\sin{3x} \ dx = \frac{-1}{3}e^{2x}cos(3x) + \frac{2}{9}e^{2x}sin(3x) - \frac{4}{9}\int e^{2x}sin(3x) dx$

$\int e^{2x}\sin{3x} \ dx + \frac{4}{9}\int e^{2x}sin(3x) dx = \frac{-1}{3}e^{2x}cos(3x) + \frac{2}{9}e^{2x}sin(3x)$

$\frac{13}{9}\int e^{2x}\sin{3x} \ dx = \frac{-1}{3}e^{2x}cos(3x) + \frac{2}{9}e^{2x}sin(3x)$

$13\int e^{2x}\sin{3x} \ dx = -3e^{2x}cos(3x) + {2}e^{2x}sin(3x)$

$\int e^{2x}\sin{3x} \ dx = \frac{-3e^{2x}cos(3x) + {2}e^{2x}sin(3x)}{13}$

$\int e^{2x}\sin{3x} \ dx = \frac{e^{2x}}{13}(2sin(3x) - 3cos(3x)) + C$

Nagisa · « **Reply #32 on:** December 26, 2012, 02:03:18 am »

thanks man, dont know why you had to show off ur complex integration skills(wasnt needed). but ye i wasnt movin them fractions around enough. TY BEBCAKZ

Hancock · « **Reply #33 on:** December 26, 2012, 02:07:26 am »

Personally I find it heaps quicker and easier using the complex exponential than double int by parts; hence why I did it that way first.

Sozza b^3

b^3 · « **Reply #34 on:** December 26, 2012, 02:08:39 am »

Quote from: Hancock on December 26, 2012, 02:07:26 am

Personally I find it heaps quicker and easier using the complex exponential than double int by parts; hence why I did it that way first.

Sozza b^3

Turns out I integrated e^(2x)*cos(3x) instead of e^(2x)*sin(3x) anyways

.... I need sleep..

Nagisa · « **Reply #35 on:** December 26, 2012, 07:24:42 pm »

evaluate the integral $\int \tan^{3}x \sec x \ dx$

b^3 · « **Reply #36 on:** December 26, 2012, 07:43:06 pm »

$\begin{alignedat}{1}\int\tan^{3}\left(x\right)\sec\left(x\right)dx & =\int\tan\left(x\right)\left(\sec^{2}\left(x\right)-1\right)\sec\left(x\right)dx \\ \mathrm{Let\:}u=\sec\left(x\right) \\ \frac{du}{dx} & =\tan(x)\sec(x) \\ \int\tan\left(x\right)\left(\sec^{2}\left(x\right)-1\right)\sec\left(x\right)dx & =\int\left(u^{2}-1\right)\tan(x)\sec(x)\frac{du}{\tan(x)\sec(x)} \\ & =\int u^{2}-1du \\ & =\frac{1}{3}u^{3}-u+C \\ & =\frac{1}{3}\sec^{3}\left(x\right)-\sec\left(x\right)+C \end{alignedat} $

EDIT: It's probably not best to show the $\frac{du}{\tan(x)\sec(x)}$ line but I've put it in there to show how we canceled the other term out when we change the variables, also the first step comes from the fact that $\tan^{2}(x)+1=\sec^{2}(x)$

Nagisa · « **Reply #37 on:** December 26, 2012, 08:00:53 pm »

thx battman, i got same thing. and the line to help me understand ur working was nice thankyou

Nagisa · « **Reply #38 on:** December 26, 2012, 08:04:31 pm »

now im havin trouble with $\int_{0}^{\frac{\pi}{2}} \sin^2x \ cos^2x \ dx$ not sure wat to do when substituting.

b^3 · « **Reply #39 on:** December 26, 2012, 08:09:35 pm »

Try using the double angle formula for cos, $\cos(2x)=1-2\sin^{2}(x)=2\cos^{2}(x)-1$ so that you get
$\sin^{2}(x)=\frac{1-\cos(2x)}{2}$ and $\cos^{2}(x)=\frac{1+\cos(2x)}{2}$ , from that you should be able to manipulate it again so that you get another use of the tule above except with $2x$ and $4x$ .

Hope that helps.

polar · « **Reply #40 on:** December 26, 2012, 08:21:11 pm »

Spoiler

$\begin{aligned} \int_{0}^{\frac{\pi}{2}}(\sin^2(x)\cos^2(x))dx &= \int_{0}^{\frac{\pi}{2}}(\sin(x)\cos(x))^2 dx \\ &= \int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}\times2\sin(x)\cos(x)\right)^2 dx \\ &= \int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}\times\sin(2x)\right)^2 dx \\ &= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}\sin^2(2x)dx \\ &= \frac{1}{4}\times\frac{1}{2}\int_{0}^{\frac{\pi}{2}} 1 - \cos(4x) dx \\ &= \frac{1}{8}\left[ x - \frac{1}{4}\sin(4x)\right]_{0}^{\frac{\pi}{2}} \\ &= \frac{1}{8}\left( \frac{\pi}{2} - \frac{1}{4}\sin\left(4\times\frac{\pi}{2}\right) - 0 + \frac{1}{4}\sin(4\times0)\right) \\ &= \frac{\pi}{16}\end{aligned}$

hopefully I've done all the brackets right this time

Nagisa · « **Reply #41 on:** December 26, 2012, 08:38:51 pm »

Quote from: polar on December 26, 2012, 08:21:11 pm

hopefully I've done all the brackets right this time

hahaha ur a legend polar ty once again

Nagisa · « **Reply #42 on:** December 27, 2012, 12:04:16 am »

Evaluate $\int \tan^5x \ dx$

$\int \tan^5x \ dx = \int \tan x \tan^2x \tan^2x \ dx = \int \tan x \ (\sec^2 - 1) (\sec^2 - 1) \ dx$

$\int \tan x \ (\sec^4x - 2\sec^2x + 1) \ dx$

$\int \tan x \sec^4x - \tan x \ 2\sec^2x + \tan x \ dx$

$\int (\tan x \sec x)(\sec^3x) - (\tan x \sec x)(2\sec x) + \tan x \ dx$

$let \ u = \sec x \ \ \ \ \ \ \ so \ du = \sec x \tan x \ dx$

$\int u^3 - 2u + \tan x \ du$

$\left [\frac{u^4}{4} - u^2 + \ln |\sec x | + C\right ]$

$\frac{1}{4}\sec^4x - \sec^2x + \ln |\sec x | + C$

it seems correct but im not sure of myself lol can anyone check it or post their opinion?

b^3 · « **Reply #43 on:** December 27, 2012, 12:08:11 am »

The line
$\int u^{3}-2u+\tan(x) du$
isn't entirely correct since you have an $x$ and you integrated with respect to $u$ and you canceled a $\sec(x)\tan(x)$ out that wasn't there for the $\tan(x)$ . What you should have done is to split the integral up, work on the first part and cancel the $\sec(x)\tan(x)$ out and then just integrate the $\tan(x)$ on it's own.

Nagisa · « **Reply #44 on:** December 27, 2012, 12:15:56 am »

Quote from: Indian Battman on December 27, 2012, 12:08:11 am

The line
$\int u^{3}-2u+\tan(x) du$
isn't entirely correct since you have an $x$ and you integrated with respect to $u$ and you canceled a $\sec(x)\tan(x)$ out that wasn't there for the $\tan(x)$ . What you should have done is to split the integral up, work on the first part and cancel the $\sec(x)\tan(x)$ out and then just integrate the $\tan(x)$ on it's own.

yea i see, thankyou man

Search the forums now!

We have moved!

Author Topic: Nagisa Maths Thread (Read 21320 times) Tweet Share

Nagisa

Re: Nagisa Maths Thread

Hancock

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

Hancock

Re: Nagisa Maths Thread

b^3

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

b^3

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

b^3

Re: Nagisa Maths Thread

polar

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

b^3

Re: Nagisa Maths Thread

Nagisa

Re: Nagisa Maths Thread

Recent Posts

User:
Password: