Need help with Integrals

[burnt compass]

Could someone explain for me how to solve this integral with trigonometric substitution? There's a lot of videos in internet, but none of them explain why they're using the things that they do

[frozen pebble]

The most common approach here is to let x = tan(theta)

The reason behind this is that the square root will turn into just sec(theta)

Thanks to the identity 1 + tan^2 = sec^2

[burnt compass]

aaaaaahhhhhhhh

makes all sense

got it

but

sorry if i sound basic here, but what does theta has to do with x?

all the materials that i've studied used it, but i don't understand the relation between them

[frozen pebble]

Theta is the new variable we are introducing, the relationship between theta and x is up to us

[burnt compass]

so it's like

we have x = tangent theta

got it

but what is this theta?

and after all the integration, how can we use it in a equation?

[burnt compass]

(in a case of defined integral)

[frozen pebble]

After you integrate it in theta you rewrite the result with x

E.g. If you got tan^2(theta) as the antiderivative, then you rewrite that as x^2

[frozen pebble]

You have done u-sub before, right?

[burnt compass]

yes

i understand all the concept

[frozen pebble]

Well you can think of this as theta = arctan(x)

[burnt compass]

O

wait

i think i got it

so you basically just attribute x = tangent theta

like you said

OOOOOOO

i'm so sorry now i got it

it's so simple

you literally just re-write x with another value

that you can integrate

got it

and now?

[frozen pebble]

Hm? What now?

[burnt compass]

yes

the rest of the integral i mean

[frozen pebble]

Talking about the problem that you posted?

[burnt compass]

yes

[frozen pebble]

Alright

So you do x = tan(theta), dx = sec^2(theta)dtheta

[burnt compass]

the first question i made was just a random one i had, sorry

[burnt compass]

rgt

[frozen pebble]

So this becomes $\int\sqrt{\tan^2\theta+1}\sec^2\theta\dd{\theta}$ or simply $\int\sec^3\theta\dd{\theta}$

[lusty harbor]

A Lonely Bean

[burnt compass]

now we can simply solve it

wow it was way easier than i thought it was

tysm bro

[frozen pebble]

Yeah, another method for solving this would be to instead do x = sinh(t) assuming you are familiar with hyperbolic trig functions

[burnt compass]

hmm

i know it, but i don't know how dx would be

.solved