Question

∫ sec^5x dx

Answer 1

@ganeshie8

Answer 2

@RadEn

Answer 3

@surjithayer

Answer 4

@aaronq

Answer 5

@phi

Answer 6

@madrockz

Answer 7

@ganeshie8 help sir

Answer 8

@LastDayWork

Answer 9

"@ganeshie8 help sir " o.O
I thought she's a girl..

Answer 10

I don't know..she is a girl? then i should say mam :)

Answer 11

help @LastDayWork :)

Answer 12

Er..you shouldn't tag me when I am already here..it causes an unnecessary notification..
Like @FaizanHBO_Channel ;)

Answer 13

haha :D

Answer 14

Talking about your question..
Separate it into
sec^4(x) * sec(x)

Can you tell me the next step ??

Answer 15

(1+tan^2x)^2* secx?

Answer 16

Correct :)
Now tell me a favorable substitution..

Answer 17

should i let tanx=z?

Answer 18

or sec x =z?

Answer 19

What do you think will create less clutter ??

Answer 20

Think in terms of substitution dx

Answer 21

*substitution of dx

Answer 22

d/dx secx=dz/dx
=>secx tanx dx=dz
or d/dx tanx=dz/dx
=>sex^2x dx=dz ??

Answer 23

I have to admit..I didn't see that one coming..just gimme a min..

Answer 24

sure :)

Answer 25

\[letI=\int\limits \sec ^{5}x~dx=\int\limits \sec ^{3}x \sec ^{2}x~dx\]
\[=\sec ^{3}x ~\tan x-\int\limits \left( 3\sec ^{2}x \sec x \tan x \right)\tan x~dx\]
\[=\sec ^{3}x \tan x-3\int\limits \sec ^{3}x \tan ^{2}x~dx\]
\[=\sec ^{3}x \tan x-3 \int\limits \sec ^{3}x \left( \sec ^{2}x-1 \right)dx\]
\[=\sec ^{3}x \tan x-3I+3\int\limits \sec ^{3}x~dx\]
\[4I=\sec ^{3}x \tan x+3\int\limits \sec x \sec ^{2}x~dx+c\]
again integrate by parts,
if any problem i will help.

Answer 26

okay thank you :)

Answer 27

Making a reduction formula (like surjithayer did) is the only way to solve this question..

Answer 28

i know :) i am getting it :)

Answer 29

so i have to let secx-U and sec^2x=V?

Answer 30

@surjithayer please tell me the next steps...

Answer 31

\[I1=\int\limits \sec x \sec ^{2}x~dx=\sec x \tan x -\int\limits \sec x \tan x \tan x~dx\]
\[=\sec x \tan x-\int\limits \sec x \tan ^{2}x~dx\]
\[=\sec x \tan x-\int\limits \sec x \left( \sec ^{2} x-1\right)dx\]
\[=\sec x \tan x-I1+\int\limits \sec x~dx\]
\[2I1=\sec x \tan x+\ln \left| \sec x+\tan x \right|\]
now you can complete.

Answer 32

\[I1=\frac{ 1 }{ 2 }\sec x \tan x+\frac{ 1 }{ 2 }\ln \left| \sec x+\tan x \right|\]

Answer 33

\[4I=\sec ^{3}x \tan x+\frac{ 3 }{ 2 }\sec x \tan x+\frac{ 3 }{ 2 }\ln \left| \sec x+\tan x \right|+c\]
\[I=\frac{ 1 }{ 4 }\sec ^{3}x \tan x +\frac{ 3 }{ 8 }\sec x \tan x+\frac{ 3 }{ 8 }\ln \left| \sec x+\tan x \right|+C\]