Question

Make a substitution that converts the integrand to a rational function

Answer 1

\[\int\limits_{}^{}\frac{7\cos\theta}{\sin^2\theta+5\sin\theta-6}~d\theta\]

Answer 2

heEdwardsFamilytay oesn'tday nowkay oway otay peaksay igay atinalay @TheEdwardsFamily

Answer 3

@Kainui @dan815

Answer 4

@freckles

Answer 5

tan(theta/2)=x is the first sub that comes to my mind
but let me think a little more and see if we can find an easier way

Answer 6

\[\sin^2(\theta)+5 \sin(\theta)-6 \\ =(\sin(\theta)+6)(\sin(\theta)-1) \\ \frac{\cos(\theta)}{(\sin(\theta)+6)(\sin(\theta)-1)}\]
let u=sin(theta)
du=cos(theta) d theta

Answer 7

it looks like it is tons easier to go this route
and then do partial fractions

Answer 8

OHHH

Answer 9

I get it now :D

Answer 10

\[\int\limits_{}^{}\frac{du}{(u+6)(u-1)} \\ \frac{A}{u+6}+\frac{B}{u-1}=\frac{1}{(u+6)(u-1)}\]

Answer 11

cool stuff

Answer 12

Thank you!!!
I thought the question was asking me something else lol

Answer 13

I didn't put the 7 on top
you can do that though

Answer 14

well yes but with the sub it will be 7 du

Answer 15

on top

Answer 16

Ok, makes sense, you already did the sub part
well thanks again @freckles :)

Answer 17

let me know if you need further help on the problem

Answer 18

\[\frac{ A }{ u+6 }+\frac{ B }{ u-1 }\]
\[7=A(u-1)+B(u+6)\]
\(0=A+b\)
\(7=B-A\)

\(A=\large \frac{-7}{2}\) and \(B=\large\frac{7}{2}\)
\[\int\limits_{}^{}\frac{\frac{-7}{2}}{u+6}+\frac{\frac{7}{2}}{u-1} du\] \[=\frac{7}{2}\ln|u-1|-\frac{7}{2}\ln|u+6|+c\]
\(u=sin\theta\)
\[=\frac{7}{2}\ln|\sin\theta-1|-\frac{7}{2}\ln|\sin\theta+6|+c\]

Answer 19

Is this correct? >.<

Answer 20

Au+Bu+6B-A=7
A+B=0
-A+6B=7

first equation gives -A=B
B+6B=7
7B=7
B=1
So A=-1.

Answer 21

Oh yeah, there was a 6B.-.

Answer 22

Thank you @myininaya

Answer 23

np