indefinite integral of sqrt(x^2-1)

[Whale_forever]

hello, i had this problem,
somehow i forgot how to integral sqrt(x^2 -1)
thanks for the answer

 

[Deleted member 4993]

hello, i had this problem,
somehow i forgot how to integral sqrt(x^2 -1)
thanks for the answer

\(\displaystyle \int \sqrt {x^2 \, - \, 1} \, dx\)

substitute

\(\displaystyle x \, = \sec\theta\)

Now continue....

 

[Whale_forever]

uh, i got stuck at integral of (sec^3 t - sec t ) dt
this what i did:
= ? sqrt (sec^2 t -1) d (sec t)
= ? tan t tan t sec t dt
= ? (sec^2 - 1) sec t dt
= ? sec^3 t - sec t dt
then?
i do this :
= ? sqrt (tan^2 t - 1) d (tan t) <--- it seems like the beginning, where did i do wrong??? it seems so confusing....

as stupid i might be this quotient as simple as:
? sqrt(x^2 + a^2) dx = x sqrt (x^2 + a^2) - ? x^2 / sqrt (x^2+a^2) dx
= (x sqrt (x^2 + a^2))/2 +a^2 /2 ln (x + sqrt ( x^2 + a^2 ))

no need to use sec , it only made it more difficult,

thanks but no thanks to someone at upper post, and thanks to whoever made math
---- it cost me a full day only to figure the answer as simple as this ,math is tiring

 

[Deleted member 4993]

uh, i got stuck at integral of (sec^3 t - sec t ) dt
this what i did:
= ? sqrt (sec^2 t -1) d (sec t)
= ? tan t tan t sec t dt
= ? (sec^2 - 1) sec t dt
= ? sec^3 t - sec t dt
then?
i do this :
= ? sqrt (tan^2 t - 1) d (tan t) <--- it seems like the beginning, where did i do wrong??? it seems so confusing....

as stupid i might be this quotient as simple as:
? sqrt(x^2 + a^2) dx = x sqrt (x^2 + a^2) - ? x^2 / sqrt (x^2+a^2) dx
= (x sqrt (x^2 + a^2))/2 +a^2 /2 ln (x + sqrt ( x^2 + a^2 ))

no need to use sec , it only made it more difficult, --- did it though - difficulty is only in the eye of the beholder (or inept)

There are many ways to skin a squirrel:

\(\displaystyle \int \frac{\sin^2\theta}{\cos^3\theta} \, d\theta\)

\(\displaystyle = \,\int \sin\theta\cdot\frac{\sin\theta}{\cos^3\theta} \, d\theta\)

\(\displaystyle = \,(\sin\theta)\cdot(\frac{1}{2\cos^2\theta}) - \int \cos\theta\cdot\frac{1}{2\cos^2\theta} \, d\theta\)

\(\displaystyle = \, \frac{1}{2}\cdot[\sin\theta\cdot\sec^2\theta \, - \, ln(|\sec\theta + \tan\theta|)]\)

....now get used to it.

thanks but no thanks to someone at upper post, and thanks to whoever made math
---- it cost me a full day only to figure the answer as simple as this ,math is tiring

 

[galactus]

thanks but no thanks to someone at upper post, and thanks to whoever made math
---- it cost me a full day only to figure the answer as simple as this ,math is tiring

Math is not tiring. It is a beautiful thing. Think of these integrals as little puzzles to solve. As with anything in life, the more you do the better you get.

Frankly, I like to see people, especially Americans, using their heads for something besides a place to grow hair, sports, celebrities, and pop music.

Most of society likes to keep their brains as sedentary as possible. Think of it has bench presses for the cerebrum

Just ponder it, all of the gadgets we depend on in life. i.e cellphones, computers, etc. were all developed using lots of math.

 

[galactus]

Trig sub can be handy. You have it down to the correct trig form. Now, to make things easier, just use the reduction formula:

\(\displaystyle \int sec^{n}(x)=\frac{sec^{n-1}(x)tan(x)}{n-1}+\frac{n-2}{n-1}\int sec^{n-2}(x)dx\)

Plug in your n values and fill it out and you have it.

Just something I thought I would show you.