Integrate
3^sin(x) cosx

Question

Integrate
3^sin(x)  cosx

tkhunny · Answer

Have you considered u = sin(x)?

anonymous · Answer

yes

anonymous · Answer

i guess I got stuck at $$\int\limits_{?}^{?}3^u$$

anonymous · Answer

du

tkhunny · Answer

$\int 3^{u}\;du = \dfrac{3^{u}}{ln(3)} + C$

That should be in a book, somewhere.

anonymous · Answer

replace 3^sin(x)  with e^(ln(3)*sinx)

anonymous · Answer

and then use sub:u=ln(3)*sinx

anonymous · Answer

that seems more complicated than it has to be :)
but ill try it

tkhunny · Answer

@ASAAD123 is demonstrating a good general method for all sorts of exponential things.  It is a little more complicated than it needs to be, but only visually.  If you can get past the eye pain, it should work.  Or, you can just know about exponentials with bases other than e, which is where I went.

Why do you care about the limits?  Is this definite integral and you just didn't tell us?  If so, you'll have to play with the limits, too.

anonymous · Answer

does the e^ln cancel?

anonymous · Answer

no it is easier ,$$\int\limits_{}^{}e ^{(\ln3) \times sinx}cosx  dx$$

anonymous · Answer

u=ln3 *sinx
du=ln3 *cosx  dx
................

anonymous · Answer

so no product rule to find du?

anonymous · Answer

no ,why ? ln3 is a number.

anonymous · Answer

right

anonymous · Answer

how do yall master this stuff? there's so many rules.

tkhunny · Answer

Please don't ask about cancelling.  Exponents and logarithms are functions requiring arguments.  $e^{ln}$ doesn't mean ANYTHING.

Like I said, it's fine if you can get past the visual pain.  It is more visually complicated.  There was no product rule int he first place.

Find methods that work well for you and try to use the ones with which you are most comfortable.  Also, and perhaps more importantly, keep everything in yer head and do some exploration.  Do things the easy way and the hard way and any other way you can think of.  You will gain experience.

anonymous · Answer

so my final answer is $$\frac{ 1 }{ \ln3 }e^{\ln3sinx}+c$$ +C

anonymous · Answer

yes

anonymous · Answer

or $$\frac{ 3^{sinx} }{ \ln 3 }+c$$

anonymous · Answer

I wont even ask how you got that or how it is the same as what I got lol

anonymous · Answer

oh it canceled, nvm  I see lol

anonymous · Answer

TYsoVM