Integrating sec²(t)/(1+tan(t))² using substitution

  • Context:
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Chain Chain rule
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 2K views
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$$\tiny\text{Whitman 8.7.15 Chain Rule} $$
$$\displaystyle
I=\int \frac{\sec^2\left({t}\right)}{\left(1+\tan\left({t}\right)\right)^2}\ d{t}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)}
+ C$$
$\begin{align}\displaystyle
u& = \tan\left({t}\right)&
du&= \sec^2 \left({t}\right)\ d{t} \\
\end{align}$

$\text{so.. chain rule} $

$
\displaystyle
\int {x}^{n} \ dx = \frac{{x}^{x+1}}{n+1}+C
$

$\text{rewrite and integrate} $
$$I=\int \frac{1}{\left(1+u\right)^3} d{u}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)^2}
+ C
$$

$\tiny\text
{from Surf the Nations math study group}$
🏄 🏄 🏄 🏄 🏄 🏄
 
Last edited:

Discussion

Physics news on Phys.org
karush said:
$$\tiny\text{Whitman 8.7.15 Chain Rule} $$
$$\displaystyle
I=\int \frac{\sec^2\left({t}\right)}{\left(1+\tan\left({t}\right)\right)^2}\ d{t}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)}
+ C$$
$\begin{align}\displaystyle
u& = \tan\left({t}\right)&
du&= \sec^2 \left({t}\right)\ d{t} \\
\end{align}$

$\text{so.. chain rule} $

$
\displaystyle
\int {x}^{n} \ dx = \frac{{x}^{x+1}}{n+1}+C
$

$\text{rewrite and integrate} $
$$I=\int \frac{1}{\left(1+u\right)^3} d{u}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)^2}
+ C
$$

$\tiny\text
{from Surf the Nations math study group}$
🏄 🏄 🏄 🏄 🏄 🏄

You somehow changed the integrand into the reciprocal of a cubic when it was originally the reciprocal of a quadratic...
 
The given should of been a cube. Thanks for the catch. Hope the rest is OK.