Solving by Substitution Calculus

[amberzak]

Hi.

I have gone wrong somewhere, but not sure where. THis is the question:

Find the integral by using the substitution given


So I did:


I know it's wrong because the answer should be


Any ideas?

 

[wjm11]

Hi.

I have gone wrong somewhere, but not sure where. THis is the question:

Find the integral by using the substitution given
View attachment 2003

So I did:
View attachment 2004

I know it's wrong because the answer should be
View attachment 2005

Any ideas?

Do not keep the sec in the denominator. Replace it with u and cancel: u^3/u = u^2.

 

[Deleted member 4993]

Hi.

I have gone wrong somewhere, but not sure where. THis is the question:

Find the integral by using the substitution given
View attachment 2003

So I did:
View attachment 2004

I know it's wrong because the answer should be
View attachment 2005

Any ideas?

you had \(\displaystyle \int \dfrac{u^3}{sec(x)}du\)

and you integrated that "pretending" x to be constant. Cannot do that!!

x is a function of u ← x = sec^-1(u) according to your substitution.

u = sec(x)

du = sec(x) * tan(x) dx

sec³(x)* tan(x) dx = sec²(x)* sec(x) * tan(x) dx = u² du

Now go forth and integrate......

However, I think your original integrand should be sec²(x)tan(x) dx to match your known answer.

 

[soroban]

Hello, amberzak!

\(\displaystyle \text{Find the integral by using the given substitution .}\)
. . \(\displaystyle \displaystyle\int \sec^3x\tan x\,dx \qquad u \,=\,\sec x\)

I would approach it like this . . .

We have: .\(\displaystyle \displaystyle\int\sec^2x(\sec x\tan x\,dx)\)

Let \(\displaystyle u \,=\,\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx\)

Substitute: .\(\displaystyle \displaystyle \int u^2\,du \;=\;\tfrac{1}{3}u^3 + C\)

Back-substitute: .\(\displaystyle \frac{1}{3}\sec^3x + C\)