$\LARGE \int_{0}^{\frac{3 \pi}{2}} \sin^3 x \cos^2 x dx$

i tried solving it but the answer did not match the choices given..i think i did something wrong

Question

$\LARGE \int_{0}^{\frac{3 \pi}{2}} \sin^3 x \cos^2 x dx$

i tried solving it but the answer did not match  the choices given..i think i did something wrong

lgbasallote · Answer

lemme fix the latex

blockcolder · Answer

Can you show us your solution?

lgbasallote · Answer

okay..gimme some time

mimi_x3 · Answer

Write it as:
$$\int\limits sinxsin^{2}xcos^{2}x  =>\int\limits sinx(1-\cos^{2}x)(\cos^{2}x)  $$
Might be easier now..it's a substitution..

lgbasallote · Answer

$\int (\sin ^2x)(\cos ^2 x) \sin x dx$

$\int (1 - \cos^2 x)(\cos ^2 x) \sin x dx$

$\int (\cos ^2 x - \cos ^4 x)\sin x dx$

let u = $\cos x$
du = $-\sin x dx$

$-\int (u^2 - u^4) du$

$\LARGE \frac{-\cos ^3 x}{3} + \frac{\cos ^5 x}{5}$

mimi_x3 · Answer

looks right..

lgbasallote · Answer

hmm seems i got it...guess i just needed to put it in latex :P

mimi_x3 · Answer

lol, yeah then sub in the limit

lgbasallote · Answer

$\huge \frac{-[\cos (\frac{3\pi}{2})]^3}{3} + \frac{[\cos (\frac{3\pi}{2})]^5}{5}$

blockcolder · Answer

You forgot 0. cos(0)=1. :D

lgbasallote · Answer

oh..oh yeah..silly me...no wonder i couldnt get this

lgbasallote · Answer

uhmmm hypothetical question...what's cos(3pi/2)? this is kinda ironic...considering i did a tutorial on it not so long ago :P

mimi_x3 · Answer

0 lol
you are able to use a calculator :P

lgbasallote · Answer

hmph

mimi_x3 · Answer

why hmph-ing? like me :p

lgbasallote · Answer

coz you wouldnt teach me how to do it :P

mimi_x3 · Answer

im your teacher now huh :P

lgbasallote · Answer

you said youll guide me :P it's just the last step..just cos (3pi/2)

mimi_x3 · Answer

need anymore help?

lgbasallote · Answer

i dont know...im just kind of thinking a new topic for my next tutoriall..you?

mimi_x3 · Answer

i gotta go; the next tutorial can be on the trig inverses idk