1. ## Trigonometric Substitution

Question:
$\displaystyle \int\frac{2dt}{t\sqrt{t^4+25}}$
Attempt:
$\displaystyle \int\frac{2dt}{t\sqrt{(t^2)^2+(5)^2}}$

a=5 ,$\displaystyle t^2=5tan\theta$,$\displaystyle 2tdt=5sec^2\theta$
Since there is now differentiation for $\displaystyle sec^2\theta$ can i change it to $\displaystyle tan^2\theta+1$

2. Originally Posted by mj.alawami
Question:
$\displaystyle \int\frac{2dt}{t\sqrt{t^4+25}}$
Attempt:
$\displaystyle \int\frac{2dt}{t\sqrt{(t^2)^2+(5)^2}}$

a=5 ,$\displaystyle t^2=5tan\theta$,$\displaystyle 2tdt=5sec^2\theta$
Since there is now differentiation for $\displaystyle sec^2\theta$ can i change it to $\displaystyle tan^2\theta+1$

Hello,
Your substitution works, but from your substitution, clearly $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$, find $\displaystyle dt$ in terms of $\displaystyle \theta$.
Then change your integral accroding to this substitution; you will get (with some simplification) a simple trigonometric integral.

3. Originally Posted by General
Hello,
Your substitution works, but from your substitution, clearly $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$, find $\displaystyle dt$ in terms of $\displaystyle \theta$.
Then change your integral accroding to this substitution; you will get (with some simplification) a simple trigonometric integral.
So ,
$\displaystyle t^2=5tan\theta$
$\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$
$\displaystyle dt=\sqrt{5}\,sec(\theta)$

4. Originally Posted by mj.alawami
So ,
$\displaystyle t^2=5tan\theta$
$\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$
$\displaystyle dt=\sqrt{5}\,\sqrt{sec^2(\theta)}$
No No No. In general: $\displaystyle \left( \sqrt{f(x)} \right)^{'} \neq \left( \sqrt{f'(x)} \right)$.

Try to find $\displaystyle dt$ again, and notice that $\displaystyle \sqrt{tan(\theta)} = tan^{\frac{1}{2}}(\theta)$.