a couple calc problems

[mindshift]

i cant remember how to do these, so mabye you can help me.

what is the integral of:

sin(x)(e^x) dx

sin(x)cos(x) dx


thanks alot

 

[soroban]

Hello, mindshift!

I'll just jog your memory . . .

∫ sin(x) (e^x) dx

This will require integration by parts ... twice.

Begin with: . u .= .sin(x) . . . . . dv .= .e^x dx

. . . .Then: . du .= .cos(x) dx . . . v .= .e^x

∫ sin(x) cos(x) dx

Straight substitution: . let .u = sin(x)

 

[mindshift]

heres what i got:

for ∫ sin(x) (e^x) dx :

e^x sin(x) - (e^x cos (x) + ∫ e^x sin(x) dx )

i dont think i did it right

and for ∫ sin(x) cos(x) dx:

1/2 sin² x

 

[tkhunny]

for ∫ sin(x) (e^x) dx :

e^x sin(x) - (e^x cos (x) + ∫ e^x sin(x) dx )

i dont think i did it right

It is fine, but you're not done.

∫ sin(x) (e^x) dx = e^x sin(x) - (e^x cos (x) + ∫ e^x sin(x) dx ) + C

∫ sin(x) (e^x) dx = e^x sin(x) - e^x cos (x) - ∫ e^x sin(x) dx + C

2*∫ sin(x) (e^x) dx = e^x sin(x) - e^x cos (x) + C

∫ sin(x) (e^x) dx = ½(e^x sin(x) - e^x cos (x)) + C₁

You may wish to rewrite a few other possible ways.

 

[tkhunny]

for ∫ sin(x) cos(x) dx:

1/2 sin² x

Funny, I have:
∫sin(x) cos(x) dx = -½cos²(x) + C

You tell me if that's the same answer.

(Hint: That constant of integration is NOT optional.)

Further, you could utilize sin(2*x) = 2*sin(x)*cos(x) and manage:

∫sin(x) cos(x) dx = -¼cos(2x) + C

You tell me if that's the same answer.

(Hint: That constant of integration is NOT optional.)

 

[mindshift]

ok, thank you very much