1. ## Trigonometric Substitution

Hi everyone,

I am a college student who is new to the math help forum. I would like to thank everyone in advance.

Evaluate the indefinite integral:

∫ 36/(sqrt(35-18x-81x^2) dx

I first manipulated the square root in the denominator:

∫sqrt(-81x^+18x-35)
- ∫ sqrt( x^2-(2/9)x-(35/81) ) I removed the negative sign
- ∫
sqrt(x^2-2/9-35/81 +(1/81)) Complete the square
- ∫ sqrt (x+(1/9)^2)

So, (x + (1/9)) = 2/3 tan θ
(3/2)x + 1/6= tan θ

I then used u-substitution, where u = 2/3 tan θ and du= 2/3 sec^2θ dθ
-∫ 36dθ/ u^2+(4/9)
-36 ∫ (2/3) sec^2θ/ (2/3) secθ
= -36∫ secθ
=-36 | secθtanθ| + C
θ= arctan (3/2)x+(1/6)

I have typed this into my Webwork, and yet the solution is wrong...any suggestions are welcome.

Also,
I am unsure where to start for this problem:

∫e^(7x)/(e^(14x)+36)dx

Would I use the natural log?

-Nikayla

2. Originally Posted by newman611
Hi everyone,

I am a college student who is new to the math help forum. I would like to thank everyone in advance.

Evaluate the indefinite integral:

∫ 36/(sqrt(35-18x-81x^2) dx

I first manipulated the square root in the denominator:

∫sqrt(-81x^+18x-35)
- ∫ sqrt( x^2-(2/9)x-(35/81) ) I removed the negative sign
Excuse me but how do you remove a negative sign inside a square root?
$\displaystyle \sqrt{-x}\ne -\sqrt{x}$

- ∫ sqrt(x^2-2/9-35/81 +(1/81)) Complete the square
- ∫ sqrt (x+(1/9)^2)

So, (x + (1/9)) = 2/3 tan θ
(3/2)x + 1/6= tan θ

I then used u-substitution, where u = 2/3 tan θ and du= 2/3 sec^2θ dθ
-∫ 36dθ/ u^2+(4/9)
-36 ∫ (2/3) sec^2θ/ (2/3) secθ
= -36∫ secθ
=-36 | secθtanθ| + C
θ= arctan (3/2)x+(1/6)

I have typed this into my Webwork, and yet the solution is wrong...any suggestions are welcome.

Also,
I am unsure where to start for this problem:

∫e^(7x)/(e^(14x)+36)dx

Would I use the natural log?

3. For the second problem, you have $\displaystyle \displaystyle \int \frac{e^{7x}}{e^{14x}+36}\;{dx} = \int \frac{e^{7x}}{(e^{7x})^2+36}\;{dx}$.
Let $\displaystyle u = e^{7x}$ to get $\displaystyle \displaystyle \frac{1}{7}\int\frac{1}{u^2+36}\;{du}$. Let $\displaystyle u = 6\tan{t}$ to get the answer.