Question

integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

Answer 1

Is that \(\sin(2x)\) or \(\sin^2(x)\)?

Answer 2

former

Answer 3

you workin on it?

Answer 4

This is very tricky. Notice that \([4(\sin(x)-\cos(x))]^2=16(\sin^2(x)+\cos^2(x)-\sin(2x))=16-16\sin(2x)\). So the denominator of the integreal is equal to \(25-(4(\sin x - \cos x))^2=[5-(4\sin x -4\cos x )]\cdot [5+(4\sin x-4\cos x)]\)
Now let \(u=4\sin x-4\cos x\). Then \(du=4(\sin x+\cos x)\,dx\). So the integral is equivalent to
\[\frac{1}{4}\int \frac{du}{(5-u)(5+u)}\]
Then you may continue from there :).

Answer 5

i cant get the answer..:-(

Answer 6

\(\frac{1}{(5-u)(5+u)}=\frac{1}{10}(\frac{1}{5-u}+\frac{1}{5+u})\)

Answer 7

how about \[du/5^{2}-u ^{2}=1/10 \log(u+5)/(u-5)\]

Answer 8

yes, that is correct!

Answer 9

i cant work it through... asked some other guys but had no luck

Answer 10

well then your answer is
\(\frac{1}{40}\ln|\frac{4\sin x-4\cos x+5}{4\sin x-4\cos x-5}|\)
and then you plug in \(x=\pi/2\) and \(x=0\).

Answer 11

my book says the answer is 1/10 log3

Answer 12

\(\frac{1}{40}(\ln(9)-\ln(1/9))=\frac{1}{40}(4\ln(3))=\frac{1}{10}\ln 3\)

Answer 13

just a little simplification would hav done it..
thanx a lot mate

Answer 14

Seriously, watchmath, where did you even think to rewrite it like that? I understand how to do the partial fractions/whatever. But I would never think to come up with something like the to rewrite it o.o

Answer 15

he has got a pretty superior brain i guess...lol