Question

integrate (3sinx-7)cosxdx/4sin^2x+9

Answer 1

sin x = t

ull get

(3t-7)dt/ 4t^2 + 9

Answer 2

now rewrite

3t - 7 = A(8t) + B

Answer 3

this is integration by parts

Answer 4

no

Answer 5

you will get a ln and an inverse tan

Answer 6

i cant understand.. sorry

Answer 7

see first substitute

Answer 8

\[\frac{3t-7}{4t^2 +9} dt \]

Answer 9

now you split the number up so it looks the derivative of the bottom

Answer 10

thats why he did

3t-7 = A (8t) +B

Answer 11

got it?

Answer 12

A= (3/8)

B= -7

Answer 13

\[= \frac{3}{8} \frac{8t}{4t^2 +9} - \frac{7}{4t^2+9} \]

Answer 14

first fraction integrates to a natural log

Answer 15

the second to an inverse tangent

Answer 16

\[= \frac{3}{8} \ln ( 4t^2+9) - \frac{7}{3} \tan^{-1} \frac{2t}{3} +C \]

Answer 17

fairly sure thats it , then sub back for t

Answer 18

the arc tangent will look a bit messy when you sub back , but I dont think you can simplify it further

Answer 19

the arc tan should be (7/6) tan^-1(2t/3)

Answer 20

poochie, have you done partial fractions?

Answer 21

its not even partial fractions , you are just spliting up the numerator and getting into a better form