# Thread: Integral (finished--need someone to check!)

1. ## Integral (finished--need someone to check!)

the integral is from 0 to π/4:

(1 + tanθ)^3(sec^2θ)dθ

So, here's what I did:

u = 1 + tanθ
du = sec^2θdθ

so,

(u)^3du

then, I changed the integral limits because I think I'm supposed to (not sure?)

1 + tanθ(π/4) = 2
1 + tanθ(0) = 1

so,

1/4(u)^4

then I simply plugged the new limit numbers into the equation to finish it

[1/4(2)^4] - [1/4(1)^4] = 15/4

Did I do it right?

2. Originally Posted by OhhCalculus
the integral is from 0 to π/4:

(1 + tanθ)^3(sec^2θ)dθ

So, here's what I did:

u = 1 + tanθ
du = sec^2θdθ

so,

(u)^3du

then, I changed the integral limits because I think I'm supposed to (not sure?)

1 + tanθ(π/4) = 2
1 + tanθ(0) = 1

so,

1/4(u)^4

then I simply plugged the new limit numbers into the equation to finish it

[1/4(2)^4] - [1/4(1)^4] = 15/4

Did I do it right?
Seems right to me.

3. Does this give you the same answer?

$\displaystyle \left[ \frac{1}{4}(\tan \theta+1)^4\right]_0^{\frac{\pi}{4}}$

4. Originally Posted by OhhCalculus
the integral is from 0 to π/4:

(1 + tanθ)^3(sec^2θ)dθ

So, here's what I did:

u = 1 + tanθ
du = sec^2θdθ

so,

(u)^3du
Execellent!

then, I changed the integral limits because I think I'm supposed to (not sure?)
You could either
1) Do the integral in terms of u, then write the answer in terms of $\theta$ again and evaluate between the limits of the original integral.
2) change the limits of integration to values of u and then evaluate the between those limits.

In my opinion, it is better and easier to do (2) as you do here.
Of course, you do NOT use the " $\theta$" values with the "u" variable!

1 + tanθ(π/4) = 2
1 + tanθ(0) = 1

so,

1/4(u)^4

then I simply plugged the new limit numbers into the equation to finish it

[1/4(2)^4] - [1/4(1)^4] = 15/4

Did I do it right?