MHB Alternative Method for Solving ∫sin(3x)cos(5x)dx?

[paulmdrdo1]

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]

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 $x$.

 

[paulmdrdo1]

is there another way of solving this? please let me know.

 

[soroban]

Hello, paulmdrdo!

Is there another way of solving this?

. . \displaystyle \int \sin(3x)\cos(5x)\,dx

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

. . \sin(3x) \:=\:3\sin x - 4\sin^3\!x
. . \cos(5x) \:=\:5\cos x - 20\cos^3\!x + 16\cos^5\!xHence:

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

. . =\;15\sin\cos x - 60\sin x\cos^3\!x + 48\sin x\cos^5\!x
. . . . . - 20\sin^3\!x\cos x + 80\sin^3\!x\cos^3\!x - 64\sin^3\!x\cos^5\!xNow integrate that term by term.

 

[alyafey22]

$$\int \sin(ax) \cos(bx) \, dx $$

Can be transformed to

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

Try integration by parts twice .

 

[DreamWeaver]

Soroban, you a meenie! Bad mammal! (Sun)(Devil)
EDIT:

[ps. High five? (Wasntme) ]

 

[Ackbach]

$$\int \sin(ax) \cos(bx) \, dx $$

Can be transformed to

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

Try integration by parts twice .

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