DLS
  • DLS
Integrate :
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
\[\LARGE \int\limits_{0}^{\frac{\pi}{3}} \frac{\sec x \tan x }{1+\sec^2 x}dx\]
DLS
  • DLS
Attempt: Let secx=t \[\LARGE \int\limits\limits_{0}^{\frac{\pi}{3}} \frac{\cancel{\sec x \tan x }}{1+t^2 } \times \frac{dt}{\cancel{\sec x \tan x}}\] \[\LARGE => \tan^{-1} (t)=\tan^{-1}(\sec x)\]
Psymon
  • Psymon
Thats right.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DLS
  • DLS
Final answer is not :(
DLS
  • DLS
\(\LARGE \tan^{-1}(2)\) is what I'm getting.
Psymon
  • Psymon
Well your initial answer without the limits applied was correct O.o
DLS
  • DLS
but since domain of tan^{-1}x is -pi/2 to pi/2 so..we need some changes here
DLS
  • DLS
wait :|
DLS
  • DLS
\[\LARGE \tan^{-1}(\sec{\frac{\pi}{3}})-\tan^{-1}(\sec0)=>\tan^{-1}(2)-\frac{\pi}{4}\]
Psymon
  • Psymon
Yeah, thats what I get.
DLS
  • DLS
\[\LARGE \int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1+\sin x)(2+\sin x)}dx\] another problem i am stuck on :(
Psymon
  • Psymon
Did ya try partial fractions?
DLS
  • DLS
didn't work
DLS
  • DLS
Tried in a different way. Let sinx=t \[\LARGE \int\limits\limits_{0}^{1} \frac{dx}{(1+t)(2+t)}dt\]
DLS
  • DLS
now we can use partial
Psymon
  • Psymon
Yep, that gets the correct answer it looks like.
DLS
  • DLS
A=1 and B= -1 ?
Psymon
  • Psymon
Yep
DLS
  • DLS
\[\LARGE \int\limits_{0}^{1} \frac{1}{1+t}-\frac{1}{2+t}dt\]
Psymon
  • Psymon
Right.
DLS
  • DLS
\[\LARGE \log|1+t|-\log|2+t| =>\log (\frac{|1+t | } { |2+t|})\]
Psymon
  • Psymon
yep yep, now back substitute and apply limits.
DLS
  • DLS
\[\LARGE \log(\frac{2}{3})-\log(\frac{1}{2})=>\log(\frac{4}{3})\]
Psymon
  • Psymon
tada! : )
Psymon
  • Psymon
Yeah, youre awesome with these, once you get the right idea you're golden : )
DLS
  • DLS
yep :D
Psymon
  • Psymon
good job xD
DLS
  • DLS
thanks for motivating o.O
Psymon
  • Psymon
Hey, Ive seen ya work, you do a good job, absolutely :3
DLS
  • DLS
ty :)
Psymon
  • Psymon
:3

Looking for something else?

Not the answer you are looking for? Search for more explanations.