# Thread: Integrating 1 + 16x^2

1. ## Integrating 1 + 16x^2

Evaluate the integration below:

2. Have you tried a trig sub?

3. Originally Posted by pickslides
Have you tried a trig sub?
yes i have tried it by using 4x = tan a but i m stuck at the last stage of integraiton, so plz help me

4. What about $\displaystyle x = \frac{1}{4}\tan u$

5. Hello, soni!

$\displaystyle \text{Evaluate: }\;\int^1_0\sqrt{1+16x^2}\,dx$

Let $4x \,=\,\tan\theta \quad\Rightarrow\quad dx \:=\:\frac{1}{4}\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{1+16x^2} \:=\:\sec\theta$

$\displaystyle\text{Substitute: }\:\int \sec\theta\cdot\tfrac{1}{4}\sec^2\!\theta\,d\theta \;=\;\tfrac{1}{4}\int \sec^3\!\theta\,d\theta$

And we have the dreaded secant-cubed integral . . .

The formula can be derived by integrating-by-parts twice.

Or we can simply memorize the formula:
. . $\displaystyle \int\sec^3\!\alpha\,dx \:=\:\tfrac{1}{2}\bigg[\sec\alpha \tan\alpha + \ln\left|\sec\alpha + \tan\alpha\right|\bigg] + C$

So we have: . $\frac{1}{4}\cdot\frac{1}{2}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C$

Back-substitute: . $\tan\theta = 4x,\;\sec\theta = \sqrt{1+16x^2}$

And we have: . $\frac{1}{8}\bigg[4x\sqrt{1+16x^2} + \ln\left|\sqrt{1+16x^2} + 4x\right|\bigg]^1_0$

Got it?

6. What I would have done after reaching $\frac{1}{4}\int sec^3(\theta)d\theta$

Rewrite it as $\frac{1}4}\int\frac{d\theta}{cos^3(\theta)}$

Since that is now an odd power of $cos(\theta)$, multiply both numerator and denominator by $cos(\theta)$ to get

$\frac{1}{4}\int\frac{cos(\theta)d\theta}{cos^4(\th eta)}= \frac{1}{4}\int\frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}$
and make the substitution $u= sin(\theta)$:

$\frac{1}{4}\int\frac{du}{(1- u^2)^2}= \frac{1}{4}\int\frac{du}{(1- u)^2(1+ u)^2}$
a rational integral. It can be done by partial fractions:
$\frac{1}{(1- u)^2(1+ u)^2}= \frac{A}{1- u}+ \frac{B}{(1- u)^2}+ \frac{C}{1+ u}+ \frac{D}{(1+ u)^2}$
Eventually giving the same result Soroban got.