Question

What is the integral of [(tanx)^3(secx)^6]dx ?

Answer 1

\[\large\rm \int\limits_{ }^{ } (tanx)^3 (secx)^6 dx\]

Answer 2

option: break sec^6x nto sec^2x
or
option 2: rewrite tan^3x as tan^1x

Answer 3

\[\huge\rm \int\limits_{ }^{} tanx \cdot \tan^2x \cdot ( secx)^6 dx\]
let u =secx

Answer 4

\[\large\rm u = \sec x , ~~du=secx ~tanx~ dx\]u=secx
so we can replace (secx)^6 with u\[\large\rm \int\limits_{ }^{ } \tan x \cdot \tan^2x \cdot u^6 \frac{du}{secx~tanx}\]
remember the identity 1+tan^2x=sec^2x
solve for tan^2x we will get `tan^2x =sec^2x-1` plug in for tan^2x
u^6 can be written as u times u^5 right ?
\[\large\rm \int\limits_{ }^{ } \tan x \cdot \tan^2x \cdot u \cdot u^5 \frac{du}{secx~tanx}\]
replace u =secx
\[\large\rm \int\limits_{ }^{ } \tan x \cdot \color{Red}{\tan^2x} \cdot (secx)\cdot u^5 \frac{du}{secx~tanx}\]
\[\large\rm \int\limits_{ }^{ } \tan x \cdot \color{Red}{(sec^2x-1)} \cdot (secx)\cdot u^5 \frac{du}{secx~tanx}\]
\[\large\rm \int\limits_{ }^{ } \tan x\cdot (secx) \cdot \color{Red}{(sec^2x-1)} \cdot u^5 \frac{du}{secx~tanx}\]
now we cancel out some *stuffs* make sense or no ??

Answer 5

can*

Answer 6

\[\frac{ (secx)^8 }{ 8 }-\frac{ (secx)^6 }{ 6 } +C\]

Answer 7

correct

Answer 8

Is that the answer?

Answer 9

okay. Thanks

Answer 10

yw