Integration help (1 Viewer)

[blackops23]

Hi guys, how would you go about integrating these?

Q1. 1/(1+(sinx)^2)

Q2. ((sinx)^2)/(1+(sinx)^2)

Q3. cosx/(1+(sinx)^2)

Any hints or tips that do NOT involve the t-formula would be extremely helpful

 

[ohexploitable]

Q3. let u=sinx

 

[AAEldar]

Question 3:

 

[ohexploitable]

for q1 and q2 i think t-sub would be best

 

[stampede]

q1 i cant be bothered figuring out, some1 else will



q2 change to :

then simplify that to

then do it from q1


q3 let u = sinx

 

[deterministic]

Question 1

Then let So:

which should be easy to solve

 

[blackops23]

Question 1

Then let So:

which should be easy to solve

pure genius, wouldn't have seen that in a million years, thank you very much.

 

[blackops23]

Ok, problem, I actually modified and simplified the problem just to get some hints into how I'd solve my question,

the real question was:

INTEGRATE: 1/(3cos^2 x + 4sin^2 x) 0(<)x(<)pi/2

which = INT 1/(3+sin^2 x) 0(<)x(<)pi/2

Following deterministic's advice:

= INT (sec^2 x)/(3sec^2 x + tan^2 x) 0(<)x(<)pi/2

= INT (sec^2 x)/(3 + 4tan^2 x) 0(<)x(<)pi/2

Now if u=tan(x), what will be the domain? I mean tan (pi/2) is undefined...

Anyother ways to do this?

Thanks, appreciate the help

 

[muzeikchun852]

Ok, problem, I actually modified and simplified the problem just to get some hints into how I'd solve my question,

the real question was:

INTEGRATE: 1/(3cos^2 x + 4sin^2 x) 0(<)x(<)pi/2

which = INT 1/(3+sin^2 x) 0(<)x(<)pi/2

Following deterministic's advice:

= INT (sec^2 x)/(3sec^2 x + tan^2 x) 0(<)x(<)pi/2

= INT (sec^2 x)/(3 + 4tan^2 x) 0(<)x(<)pi/2

Now if u=tan(x), what will be the domain? I mean tan (pi/2) is undefined...

Anyother ways to do this?

Thanks, appreciate the help

im not sure whether im right but hopefully it helps.

 

[deterministic]

Upper bound would be infinity

Note
So the result should be:

Someone else can confirm this.

 

[muzeikchun852]

Upper bound would be infinity

Note
So the result should be:

Someone else can confirm this.

yep. i think it's right.

 

[blackops23]

Sorry, how would the upper bound be infinity?

I mean u=tan(x), so when x=pi/2, u=undefined.... isn't it?

I mean its not u=inverse tan x, so inverse tan (pi/2) = infinity...?

Can someone please offer an explanation?

Thanks

 

[Hermes1]

Sorry, how would the upper bound be infinity?

I mean u=tan(x), so when x=pi/2, u=undefined.... isn't it?

I mean its not u=inverse tan x, so inverse tan (pi/2) = infinity...?

Can someone please offer an explanation?

Thanks

you let u = tanx

on the graph of y = tan x
as x approaches pi/2
tanx approaches infinity

therefore you let your upper bound limit be infinity
when you solve the integral
you get tan inverse of infinity, now on the graph of tan inverse x, as x approaches infinity it approaches pi/2

therefore you get from deterministic's answer as x approaches infinity
area under curve approaches 2/4root3 pi/2
but because infinity dominates you just write the answer