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Integration of trig. function.

paulmdrdo

Active member
May 13, 2013
386
another trig problem that i tried to solve. just want know an alternative way of solving this without using product formula.

$\displaystyle\int \sin(3x)\cos(5x)dx$

anyways this is how i solved it

$\displaystyle\int\frac{1}{2}\sin(3x-5x)+\frac{1}{2}\sin(3x+5x)dx$

$\displaystyle\int\frac{1}{2}\sin(-2x)+\frac{1}{2}\sin(8x)dx$

$\displaystyle\frac{1}{2}\int\sin(-2x)dx+\frac{1}{2}\int\sin(8x)dx$

$\displaystyle u=-2x$; $\displaystyle du=-2dx$; $\displaystyle dx=-\frac{1}{2}du$

$\displaystyle v=8x$; $\displaystyle dv=8dx$; $\displaystyle dx=\frac{1}{8}dv$

$\displaystyle-\frac{1}{4}\int\sin(u)du+\frac{1}{16}\int\sin(v)dv$

$\displaystyle -\frac{1}{4}(-\cos(-2x))+\frac{1}{16}(-\cos(8x))+C$

$\displaystyle \frac{1}{4}\cos(-2x)-\frac{1}{16}\cos(8x)+C$ ---- this is my answer.

the -2x part in this answer is bothering me because in my book it's not negative. please tell me why is that.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The $-2x$ part in this answer is bothering me because in my book it's not negative. Please tell me why is that.
Because $\cos$ is an even function. That is, $\cos(-x)= \cos(x)$ for all $x$.
 

paulmdrdo

Active member
May 13, 2013
386
is there another way of solving this? please let me know.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, paulmdrdo!

Is there another way of solving this?

. . [tex]\displaystyle \int \sin(3x)\cos(5x)\,dx[/tex]

There is . . . but you won't like it!


Identities:

. . [tex]\sin(3x) \:=\:3\sin x - 4\sin^3\!x[/tex]
. . [tex]\cos(5x) \:=\:5\cos x - 20\cos^3\!x + 16\cos^5\!x[/tex]


Hence:

.[tex]\sin(3x)\cos(5x) \,=\, (3\sin x\!-\!4\sin^3\!x)(5\cos x\!-\!20\cos^3\!x\!+\!16\cos^5\!x)[/tex]

. . [tex]=\;15\sin\cos x - 60\sin x\cos^3\!x + 48\sin x\cos^5\!x[/tex]
. . . . . [tex]- 20\sin^3\!x\cos x + 80\sin^3\!x\cos^3\!x - 64\sin^3\!x\cos^5\!x [/tex]


Now integrate that term by term.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \int \sin(ax) \cos(bx) \, dx \)

Can be transformed to

\(\displaystyle \frac{1}{a}\int \sin(x) \cos(c x) \, dx \)

Try integration by parts twice .
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Soroban, you a meenie! Bad mammal! (Sun)(Devil)



EDIT:

[ps. High five??? (Wasntme) ]
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
\(\displaystyle \int \sin(ax) \cos(bx) \, dx \)

Can be transformed to

\(\displaystyle \frac{1}{a}\int \sin(x) \cos(c x) \, dx \)

Try integration by parts twice .
You wouldn't even need to transform. Just integrate by parts twice and "solve" for the integral.