Author Topic: integration involving trigonometric identities (Read 1882 times) Tweet Share

bubbles · « **on:** July 02, 2008, 06:20:47 pm »

Please help me with these following questions:
[Ok this may sound a little stupid but is there a quick and easy way to solve these type of questions or to recognise which identity to use (im really struggling with specialist especially since i didn't do general maths advanced last yr as that subject doesn't exist at our school

and in addition we're flying through the course so im pretty much lost ><")]

Find an antiderivative:
1. tan ³x dx
2. 7cos^7 t
3. sin²xcos^4 x
4. cosec²(x- π/2)

This is fustrating me- how do i type those mathematical symbols i.e. integral, pies etc in a post?

ell · « **Reply #1 on:** July 02, 2008, 07:04:08 pm »

Quote from: bubbles on July 02, 2008, 06:20:47 pm

Please help me with these following questions:
[Ok this may sound a little stupid but is there a quick and easy way to solve these type of questions or to recognise which identity to use (im really struggling with specialist especially since i didn't do general maths advanced last yr as that subject doesn't exist at our school and in addition we're flying through the course so im pretty much lost ><")]

Find an antiderivative:
1. tan ³x dx
2. 7cos^7 t
3. sin²xcos^4 x
4. cosec²(x- π/2)

This is fustrating me- how do i type those mathematical symbols i.e. integral, pies etc in a post?

To get the fancy looking math symbols have a read of this post: http://vcenotes.com/forum/index.php/topic,3137.0.html

Generally if its an odd power, you'll need to split it up. You'll use $\sin^2 x + \cos^2 x = 1$ when it's odd (usually).

When it's just something like $\int \sin^2 x\ dx$ (an even power) you'll need to use your double-angle formulas, i.e. $\int \frac{1}{2}(1-\cos 2x)dx$

for 1.

$\int \tan^{3} xdx$

$\int \frac{\sin^{3} x}{\cos^{3} x}dx$

$\int \frac{\sin^{2} x\sin x}{\cos^{3}x}dx$

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

Now if we let $u=\cos x \Longrightarrow \frac{du}{dx}=-\sin x$

Substituting $u$ and $dx$ :

$-\int \frac{1-u^{2}}{u^{3}}du$

$-\int \frac{1}{u^{3}}-\frac{1}{u}}du$

$-\left[-\frac{1}{u^{2}}-\log_e u}\right]$ ( $c=0$ )

$\frac{1}{2u^{2}}+\log_e u}$

$\frac{\sec^{2} x}{2}+\log_e (\cos x)}$ (subbing back in $\cos x$ )

humph · « **Reply #2 on:** July 02, 2008, 07:50:23 pm »

Quote from: bubbles on July 02, 2008, 06:20:47 pm

Please help me with these following questions:
[Ok this may sound a little stupid but is there a quick and easy way to solve these type of questions or to recognise which identity to use (im really struggling with specialist especially since i didn't do general maths advanced last yr as that subject doesn't exist at our school and in addition we're flying through the course so im pretty much lost ><")]

Find an antiderivative:
1. tan ³x dx
2. 7cos^7 t
3. sin²xcos^4 x
4. cosec²(x- π/2)

This is fustrating me- how do i type those mathematical symbols i.e. integral, pies etc in a post?

1) $\tan^3 x = \tan x (\sec^2 x - 1)$ , using the standard trig identity $\tan^2 x = \sec^2 x - 1$ . So
$\begin{align*} \int{\tan^3 x \: dx} & = \int{ \tan x \: \sec^2 x \: dx} - \int{\tan x \: dx} \\ & = \int{ \tan x \: \sec^2 x \: dx} - \int{\frac{\sin x}{\cos x} \: dx} \end{align*}$
Now let $u = \tan x$ , so that $\frac{du}{dx} = \sec^2 x$ . Similarly, let $v = \cos x$ , so that $\frac{dv}{dx} = - \sin x$ . Then we can write
$\begin{align*} \int{\tan^3 x \: dx} & = \int{ u \: \frac{du}{dx} \: dx} - \int{\frac{-1}{v} \: \frac{dv}{dx} \: dx} \\ & = \int{ u \: du} + \int{\frac{1}{v} \: dv} \\ & = \frac{u^2}{2} + \ln |v| + c \\ & = \frac{1}{2} \tan^2 x + \ln |\cos x | + c. \end{align*}$
As it just says an antiderivative, we can just set $c=0$ . Note that this answer is identical to the one above if we set $c = \frac{1}{2}$ , because $\frac{1}{2} \tan^2 x + \frac{1}{2} = \frac{1}{2} \sec^2 x$ by the trig identity discussed earlier.

Mao · « **Reply #3 on:** July 02, 2008, 09:46:46 pm »

Quote from: bubbles on July 02, 2008, 06:20:47 pm

7cos^7 t

$\int 7 \cos^7 t \; dt = 7\int \cos(t)\cdot (1-\sin^2(t))^3\; dt$
then make the substitution of $u=\sin(t)$

$\implies 7\int (1-u^2)^3\; du = 7\int 1-3u^2 + 3u^4 -u^6\; du$

$\implies 7\cdot \left(-\frac{u^7}{7}+\frac{3u^5}{5}-u^3+u\right)+c$

$\implies -\sin^7(t)+\frac{21\sin^5(t)}{5}-7\sin^3(t)+7\sin(t)+c$

Quote from: bubbles on July 02, 2008, 06:20:47 pm

sin²xcos^4 x

$\int \sin^2(x)\cos^4(x)\; dx=\int (\sin(x)\cos(x))^2\cos^2(x)\; dx$

$\implies \int \frac{1}{4}\sin^2(2x)\cdot \left(\frac{\cos(2x)+1}{2}\right)\;$

$\implies \frac{1}{8}\int \sin^2(2x)\cos(2x)+\sin^2(2x)\;$

$\implies \frac{1}{8} \left( \frac{\sin^3(2x)}{3\cdot 2} + \frac{1}{2}\left(x-\frac{1}{4}\sin(4x)\right)\right) + c$

$\implies \frac{\sin^3(2x)}{48}-\frac{\sin(4x)}{64}+\frac{x}{16}+c$

Quote from: bubbles on July 02, 2008, 06:20:47 pm

cosec²(x- π/2)

$\int \csc^2\left(x-\frac{\pi}{2}\right)\; dx = \int \frac{1}{\sin^2 \left(x-\frac{\pi}{2}\right)}\; dx = \int \frac{1}{(-\cos(x))^2}\; dx = \tan(x)+c$

Mao · « **Reply #4 on:** July 03, 2008, 10:39:30 am »

Quote from: bubbles on July 02, 2008, 06:20:47 pm

This is fustrating me- how do i type those mathematical symbols i.e. integral, pies etc in a post?

http://vcenotes.com/forum/index.php/topic,3137.0.html

VCE Notes > Mathematics/Science/Technology > Mathematics > Topic: New Users: LaTeX typeset in Maths boards

bubbles · « **Reply #5 on:** July 03, 2008, 10:00:05 pm »

THANK YOU SO MUCH

bubbles · « **Reply #6 on:** July 09, 2008, 04:24:59 pm »

Query:

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

$\frac{1}{8}$ $\frac{sin^{3}(2x)}{3.2}$

Hey Mao, could you explain how'd you obtain this integral. It's much quicker than the method that I was using- integration by substitution (i.e let u = sin(2x) etc etc) I hope that makes sense ><"

ell · « **Reply #7 on:** July 09, 2008, 04:30:10 pm »

He is using substitution, he just skips a lot of steps.

bubbles · « **Reply #8 on:** July 09, 2008, 04:52:01 pm »

oh haha -_-"" lol thanks ell

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