Question

integrate... 8/(tan(4x)+3) using half angle...

Answer 1

Did you try substituting \(u=\tan({4x})\)?

Answer 2

\[\tan(4x)=\frac{2\tan(2x)}{1-\tan^2(2x)}\]

Answer 3

what i did was
u=4x
z=tan(1/2 u)

Answer 4

and i end up on partial fractions

Answer 5

you arenot asking this right?
\[\int\limits \frac{8}{\tan^4x+3}dx\]

Answer 6

nope

Answer 7

Mr. Math's sub is right on...
\[\int\frac8{\tan(4x)+3}dx\]\[u=\tan(4x)\]\[du=4\sec^2(4x)dx=4(u^2+1)dx\]\[dx=\frac{du}{4(u^2+1)}\]our integral becomes\[\int\frac2{(u+3)(u^2+1)}du\]partial fractions\[\frac A{u+3}+\frac{Bu+C}{u^2+1}=\frac2{(u+3)(u^2+1)}\]\[A(u^2+1)+(Bu+C)(u+3)=2\]\[u=-3\to 10A=2\to A=\frac15\]\[A+B=0\to B=-\frac15\]\[u=0\to\frac15+3C=2\to C=\frac35\]so now we have\[\int\frac15\frac1{u+3}+\frac15\frac{-u+3}{u^2+1}du\]\[=\frac15\int\frac1{u+3}-\frac u{u^2+1}+\frac3{u^2+1}du\]\[=\frac15\ln|u+3|-\frac1{10}\ln|u^2+1|+\frac35\tan^{-1}u+C\]\[=\frac15\ln|\tan(4x)+3|-\frac1{10}\ln|\sec^2(4x)|+\frac{12x}5+C\]Don't know how to use a half-angle formula here...

Answer 8

actually we can simplify those log terms using log rules...\[\frac15(\ln|\sin(4x)+3\cos(4x)|+12x)+C\]

Answer 9

shetty prof why ask half angle for theis

Answer 10

I tried to work them in there, but couldn't do it :/

Answer 11

i stopped on partial fractions

Answer 12

hahahha

Answer 13

Well at least you have one way to do it
good luck pleasing your professor ;)

Answer 14

thanks haha