Question

is this integral correct!
\(\large \int (sinx)^3(cosx)^4dx\)
so first i wrote \(\large sinx\times(sinx)^2(cosx)^4= sinx[(cosx)^4-(cosx)^6\)
u substitution
u= cosx==> du=-sinxdx
-du=sinxdx
so \(\large \int -(u^4-u^6)du\)
\(\large \int u^6du-\int u^4du\)
\(\large \frac{u^7}{7}-\frac{u^5}{5}+C\)
\(\large \frac{(cosx)^7}{7}-\frac{(cosx)^5}{5}+C\)

Answer 1

i put it in wolfram but the answer was diffrent

Answer 2

sorry i forget one bracket haha just learning latex hehe

Answer 3

peel off a cosine, write the rest in terms of sine and us a u sub

Answer 4

which might have been what you did
don't fret too much about the wolf, there are many trig identities that it might use to give an equivalent answer

Answer 5

So my answer is fine right? i thought i did it wrong

Answer 6

idk i didn't check but i will be happy to

Answer 7

okay try^_^!
i think wolfram used other identities to get that final expression like you said hehe

Answer 8

oh no your answer is not right

Answer 9

what's the error!

Answer 10

oh hold on maybe i am mistaken

Answer 11

sorry it is right and i was wrong
you have it

Answer 12

okay! i thought my steps were fine hehe

Answer 13

Okay thanks!

Answer 14

http://www.wolframalpha.com/input/?i=%5Cint+%28sinx%29%5E3%28cosx%29%5E4
a lot of work involved here hehe

Answer 15

http://www.wolframalpha.com/input/?i=d%2Fdx%28+cos^7%28x%29%2F7++-+cos^5%28x%29%2F5+%29

Answer 16

well it should be correct lol. i was lost with what wolfram give lol
thanks guys

Answer 17

note the use of the double angle
wolf has \(\cos(2x)\) in one of the slots