Question

\[\int\limits_{}^{}\frac{ dx }{(16+9x^2)^{\frac{ 3 }{ 2 }} }\]

Answer 1

when you see the form a^2x^2 +b^2
you plug in x = (b/a) tan u

Answer 2

here, put x = (4/3) tan u

Answer 3

what is the u?

Answer 4

new variable after substitution...

Answer 5

x = (4/3) tan u
dx = ... ?

Answer 6

4/3 sec^2 du?

Answer 7

correct.
and what does denominator simplify to ?

Answer 8

4/3 sec^3du

Answer 9

du ?
4/3 ?
i just get 64 sec^3 u

Answer 10

oh ok i see

Answer 11

then where do i go from there?

Answer 12

what gets cancelled!?

Answer 13

isn't just cos u/48 remain ?
do i need to show you the simplification part ?

Answer 14

well i'm just not thinking this is right because according to the book i should be getting. \[\frac{ x }{ 16\sqrt{16+9x ^{2}} }+C\]

Answer 15

we will get to that after simplifying the answer ..... but did u get till cos u/48 ?

Answer 16

no, i guess not

Answer 17

where did you go after we got 4/3 scec^2 du.

Answer 18

you just plugged that in for x?

Answer 19

we simplified the denominator....could you ?

Answer 20

16 + 9* 16/9 tan^2 u = 16 +16 tan^2 u = 16 sec^2 u

that ^2/3 =....

Answer 21

16^2/3 = 64
(sec^2) ^ 2/3 = sec^3
now got how the denom = 64 sec^3 x ?

Answer 22

oh ok I get that

Answer 23

so, sec^2 from numerator gets cancelled with sec^3 to get 1/ sec = cos

Answer 24

Oh i forgot we changed dx

Answer 25

till which step did u totally get ?

Answer 26

so far I've gotten everything once you actually put it in the equation. we should have \[\frac{ (4/3)cosx }{64 }\]

Answer 27

which simplifies to cos u/48

Answer 28

thats cos u/48
so integrate that

Answer 29

easiest trigo integral ever :)

Answer 30

wouldn't that be -sin u/48

Answer 31

that would be +sin u/48+c
now just put u = tan inverse (3x/4)

Answer 32

would that be (cos 3x/4)/48

Answer 33

no....
(1/48)sin (tan inverse 3x+4) +c
that will not be cos

Answer 34

oh ok

Answer 35

\(\Huge \sin (\tan^{-1}A) = \dfrac{A}{\sqrt{1+A^2}}\)
here A = 3x/4

Answer 36

ok. i don't have that on my table of integrals. so i'm going to write that down

Answer 37

sure, thats inverse trigonometric relation.

Answer 38

(1/48)sin (tan inverse 3x/4) +c
= (1/48) [(3x/4) / \sqrt {1+ (3x/4)^2}] +c
if you simplify this, you will het the answer same as that given in your book :)