1. Originally Posted by roshanhero
I am sorry but shouldnot there have been
$\displaystyle \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}$
Yes.

So the integral actually becomes

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

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

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

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

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

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

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

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

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

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

Therefore

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

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

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

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

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

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

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

$\displaystyle 2x = \tan{\theta}$

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

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

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

let u = 2x and du = 2 dx. then you would get $\displaystyle \frac{1}{2}\$$\displaystyle \int{\frac{1}{1 + u^2}\,du} then you would get the same answer \displaystyle \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 \displaystyle \frac{1}{1 + 4x^2} = \displaystyle \frac{1}{1+ (2x)^2} let u = 2x and du = 2 dx. then you would get \displaystyle \frac{1}{2}\$$\displaystyle \int{\frac{1}{1 + u^2}\,du}$

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

4. Originally Posted by Prove It
That's only if you remember the rule for $\displaystyle \arctan$, 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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