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\[\int\limits_{}^{}\sec^4(2x+1)dx\]

so far ive done this--> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]

so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]

what is the next step? could somebody help me?

now let m=4

\[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]

\[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]

u =2x+1

\[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]

An alternative way to solve this problem

there are many trig identities that can help in solving this problem one way is:

\[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]

expand the function and distribute the integral

\[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]

\[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]

\[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]

ur welcome! i was expecting ur question i did that on purpose...

now there are two ways to think about this...

the other way to visualize this concept is by assuming that tan(u) =v

dv = sec^2(u)du -> du=dv/sec^(u)

\[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]

now can u see the full picture...

(v^3)/3+c

but v=tan(u)

=(1/3)tan^3(u)+c, where u = 2x+1

hope this clarifies the point...

are u there?

So u can correct the information for ur teacher, if u think this is what she/he meant...

but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...

by the way this problem can be solved in many ways...

uhmm yeah i think i have to understand the second method first. i'll have to read it once again

just write it down on a piece of paper, you'll see how easy it is...

that was the integration

the integral of tan^2(u)sec^(u)du

now when u differentiate tan u get sec

let me show u

\[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]

let
\[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]

\[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]

can u see how sec^2(u) is cancelled so u are left with

\[\frac{1}{2} \int\limits v^2dv\]

and u know the rest, or do u want me to finish it to the end...

yes tan(u)=v therefore tan^2(u)=v^2

is that what u meant? or i misunderstood?

yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?

well in the case of the question above v^4/4+c

tan^4(u)/4 +c

v^99

tan^100(u)/100+c

if u did not have the sec^2(u), then you can't do that...

that's what i meant of the function multiplied by its derivative

if i tell u to integrate sin(x)cos(x), what answer would i get?

i think i'd use -->:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[-\cos(x)+c\]

is that correct??

u mean -(1/2)cos^2(x)+c

this is the whole point always when u integrate check if the function has its derivative...

what is the integral of sin(x)

its: -cos (u) +c :D

and what result did u get for sin(x)cos(x) again?

no never do that

a rcipical of cos never gives u a minus sign be careful never make such a mistake !

\[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}\]

:o oohh, so is that the final result? +c

no that is the result of what you thought it was the result which is not the result in this case

\[\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c\]

but how did u get the 1/2 and cos^2 ?

use the u sub and check for urself

yeep thanks! i'll try it :D :D