Question

Integrate :

Answer 1

\[\LARGE \int\limits_{0}^{\frac{\pi}{3}} \frac{\sec x \tan x }{1+\sec^2 x}dx\]

Answer 2

Attempt:
Let secx=t
\[\LARGE \int\limits\limits_{0}^{\frac{\pi}{3}} \frac{\cancel{\sec x \tan x }}{1+t^2 } \times \frac{dt}{\cancel{\sec x \tan x}}\]
\[\LARGE => \tan^{-1} (t)=\tan^{-1}(\sec x)\]

Answer 3

Thats right.

Answer 4

Final answer is not :(

Answer 5

\(\LARGE \tan^{-1}(2)\) is what I'm getting.

Answer 6

Well your initial answer without the limits applied was correct O.o

Answer 7

but since domain of tan^{-1}x is -pi/2 to pi/2 so..we need some changes here

Answer 8

wait :|

Answer 9

\[\LARGE \tan^{-1}(\sec{\frac{\pi}{3}})-\tan^{-1}(\sec0)=>\tan^{-1}(2)-\frac{\pi}{4}\]

Answer 10

Yeah, thats what I get.

Answer 11

\[\LARGE \int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1+\sin x)(2+\sin x)}dx\]
another problem i am stuck on :(

Answer 12

Did ya try partial fractions?

Answer 13

didn't work

Answer 14

Tried in a different way.
Let sinx=t
\[\LARGE \int\limits\limits_{0}^{1} \frac{dx}{(1+t)(2+t)}dt\]

Answer 15

now we can use partial

Answer 16

Yep, that gets the correct answer it looks like.

Answer 17

A=1 and B= -1 ?

Answer 18

Yep

Answer 19

\[\LARGE \int\limits_{0}^{1} \frac{1}{1+t}-\frac{1}{2+t}dt\]

Answer 20

Right.

Answer 21

\[\LARGE \log|1+t|-\log|2+t| =>\log (\frac{|1+t | } { |2+t|})\]

Answer 22

yep yep, now back substitute and apply limits.

Answer 23

\[\LARGE \log(\frac{2}{3})-\log(\frac{1}{2})=>\log(\frac{4}{3})\]

Answer 24

tada! : )

Answer 25

Yeah, youre awesome with these, once you get the right idea you're golden : )

Answer 26

yep :D

Answer 27

good job xD

Answer 28

thanks for motivating o.O

Answer 29

Hey, Ive seen ya work, you do a good job, absolutely :3

Answer 30

ty :)

Answer 31

:3