Question

please integrate sinxcosx
explanation required

Answer 1

1/2 sin 2x

Answer 2

sin2x
------
2

Now integrate

Answer 3

take it from here..

Answer 4

Ishaan is a genius :D

Answer 5

FoolForMath is

Answer 6

a Fool :D

Answer 7

as a result of undoing the chain rule
\[\int f'(g(x))g'(x)dx=f(g(x))+C\]

Answer 8

[sin(x)]' = cos(x) is a good one to know

[(sin(x))^2]' = 2 sin(x) cos(x)

so as long as there is a /2 to catch that "2"; we have it

Answer 9

what is that amistre64? why are you using that here ?

Answer 10

why am i using what here?

Answer 11

you can also do this as follows:\[u=\cos(x)\]\[du=-\sin(x)dx\]therefore:\[\int\sin(x)\cos(x)dx=-\int u du=-\frac{u^2}{2}+constant\]then substitue value for u back into result.

Answer 12

Wolfram will show you the steps as well:
http://www.wolframalpha.com/input/?i=integrate+sinxcosx&lk=3

Click on "Show Steps"

Answer 13

I meant we could simply get the answer by trig manipulation ,so no need to substitute

Answer 14

thanks alot ppl

Answer 15

$$ \int \sin x \cos x dx = \frac12 \int \sin 2x dx = - \frac14 \cos 2x +C$$

so simple..