1. Originally Posted by roshanhero
I am sorry but shouldnot there have been
$

\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}
$
Yes.

So the integral actually becomes

$\int{u^3\,\frac{du}{dx}\,dx} = \int{u^3\,du}$

$= \frac{1}{4}u^4 + C$

$= \frac{1}{4}(1 + \tan{x})^4 + C$.

2. Originally Posted by Prove It
The first two have been taken care of.

for $\int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $u$ substitution. You use a trigonometric substitution.

$\int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$

$= \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.

Now use the substitution $x = \frac{1}{2}\tan{\theta}$.

Note that $\frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.

So $dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.

Therefore

$= \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$

$= \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$

$= \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$

$= \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$

$= \frac{1}{2}\int{1\,d\theta}$

$= \frac{1}{2}\theta + C$.

Now remembering that $x = \frac{1}{2}\tan{\theta}$

$2x = \tan{\theta}$

$\theta = \arctan{2x}$.

So finally our answer is $\frac{1}{2}\arctan{2x} + C$.
Actually I think doing a u-substitution for this problem would be much easier than doing a trig substitution.

start by noticing that $\frac{1}{1 + 4x^2}$ = $\frac{1}{1+ (2x)^2}$

let u = 2x and du = 2 dx. then you would get $\frac{1}{2}\$ $\int{\frac{1}{1 + u^2}\,du}$

then you would get the same answer $\frac{1}{2}\arctan{2x} + C$ with much less work.

3. Originally Posted by oblixps
Actually I think doing a u-substitution for this problem would be much easier than doing a trig substitution.

start by noticing that $\frac{1}{1 + 4x^2}$ = $\frac{1}{1+ (2x)^2}$

let u = 2x and du = 2 dx. then you would get $\frac{1}{2}\$ $\int{\frac{1}{1 + u^2}\,du}$

then you would get the same answer $\frac{1}{2}\arctan{2x} + C$ with much less work.
That's only if you remember the rule for $\arctan$, which was found from trigonometric substitution...

4. Originally Posted by Prove It
That's only if you remember the rule for $\arctan$, which was found from trigonometric substitution...
I only used the definition of the derivative of arctan(x) along with a simple u-substitution to find this integral. no need for trig sub.

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