Question

Integrate this for me please

Answer 1

Let's do a variable substitution, by the formula $\int u\,dv=uv-\int v\,du$

$I=\int\tan^2(x)\sec^4(x)dx=\int\underbrace{\tan(x)\sec^3(x)}_{u}\underbrace{\tan(x)\sec(x)dx}_{dv}\\ u=\tan(x)\sec^3(x)\\\\ du=(\sec^5(x)+3\sec^3(x)\tan^2(x))dx\\\\ du=\sec^3(x)(\underbrace{\sec^2(x)}_{\tan^2+1}+3\tan^2(x))dx\\\\ du=\sec^3(x)(4\tan^2(x)+1)dx\\\\\\ dv=\sec(x)\tan(x)dx\\\\ v=\sec(x)$

So:

$I=\tan(x)\sec^4(x)-\int\sec^4(x)(4\tan^2(x)+1)dx\\\\ I=\tan(x)\sec^4(x)-4\underbrace{\int\sec^4(x)\tan^2(x)dx}_{I}-\int\sec^4(x)dx\\\\ I=\tan(x)\sec^4(x)-4I-\int\sec^4(x)dx\\\\ 5I=\tan(x)\sec^4(x)-\underbrace{\int\sec^4(x)dx}_{I_2}$

Solving I₂ using substitution, too:

$I_2=\int\sec^4(x)dx=\int\underbrace{\sec^2(x)}_{u}\underbrace{\sec^2(x)dx}_{dv}\\\\\\ u=\sec^2(x)\\\\ du=2\sec^2(x)\tan(x)dx\\\\\\ dv=\sec^2(x)dx\\\\ v=\tan(x)$

Then:

$I_2=\tan(x)\sec^2(x)-\int 2\tan^2(x)\sec^2(x)dx\\\\ I_2=\tan(x)\sec^2(x)-2\int\tan^2(x)\sec^2(x)dx\\\\\\ y=\tan(x)\to dy=\sec^2(x)dx\to dx=\dfrac{dy}{\sec^2(x)}\\\\ \tan^2(x)\sec^2(x)dx=y^2\sec^2(x)\dfrac{dy}{\sec^2(x)}=y^2dy\\\\\\ I_2=\tan(x)\sec^2(x)-2\int y^2dy\\\\ I_2=\tan(x)\sec^2(x)-2\cdot\dfrac{y^3}{3}\\\\ I_2=\tan(x)\sec^2(x)-\frac{2}{3}\tan^3(x)$

Hence, substituting I₂ in I:

$5I=\tan(x)\sec^4(x)-I_2\\\\ 5I=\tan(x)\sec^4(x)-(\tan(x)\sec^2(x)-\frac{2}{3}\tan^3(x))\\\\ 5I=\tan(x)\sec^4(x)-\tan(x)\sec^2(x)+\frac{2}{3}\tan^3(x))\\\\ \boxed{I=\frac{1}{5}\tan(x)\sec^4(x)-\frac{1}{5}\tan(x)\sec^2(x)+\frac{2}{15}\tan^3(x)+C}$

Now, using the limits of integration in the expression E of the statement:

$E=\displaystyle\int^{\dfrac{\pi}{6}}_0\tan^2(x)\sec^4(x)dx\\\\\\ E=(\frac{1}{5}\tan(\frac{\pi}{6})\sec^4(\frac{\pi}{6})-\frac{1}{5}\tan(\frac{\pi}{6})\sec^2(\frac{\pi}{6})+\frac{2}{15}\tan^3(\frac{\pi}{6}))-\\\\ (\frac{1}{5}\tan(0)\sec^4(0)-\frac{1}{5}\tan(0)\sec^2(0)+\frac{2}{15}\tan^3(0))\\\\\\ E=(\frac{1}{5}\cdot\frac{1}{\sqrt3}\cdot(\frac{2}{\sqrt3})^4-\frac{1}{5}\cdot\frac{1}{\sqrt3}\cdot(\frac{2}{\sqrt3})^2+\frac{2}{15}(\frac{1}{\sqrt3})^3)-\\\\ (\frac{1}{5}\cdot0\cdot1^4-\frac{1}{5}\cdot0\cdot1^2+\frac{2}{15}\cdot0^3)$


$E=\frac{1}{5\sqrt3}\cdot\frac{16}{9}-\frac{1}{5\sqrt3}\cdot\frac{4}{3}+\frac{2}{15}\cdot\frac{1}{3\sqrt3}-0+0-0\\\\\ E=\frac{1}{5\sqrt3}(\frac{16}{9}-\frac{4}{3}+\frac{2}{9})\\\\ E=\frac{1}{5\sqrt3}\cdot\frac{16-12+2}{9}=\frac{1}{5\sqrt3}\cdot\frac{6}{9}=\frac{1}{5\sqrt3}\cdot\frac{2}{3}\\\\ \boxed{E=\dfrac{2}{15\sqrt3}}$

Answer 2

$\int\limits_{0}^{\frac{\pi }{6}}tan^2(x)sec^2(x)\cdot dx \\------------------\\ u=tan(x)\implies \frac{du}{dx}=sec^2(x)\implies \frac{du}{sec^2(x)}=dx \\------------------\\ \int\limits_{0}^{\frac{\pi }{6}}u^2sec^2(x)\cdot \cfrac{du}{sec^2(x)}\implies \int\limits_{0}^{\frac{\pi }{6}}u^2\cdot du \\ \quad$

$\textit{now, we need to change the bounds as well, so} \\------------------\\ u(0)=tan(0)\implies 0 \\ \quad \\ u\left( \frac{\pi }{6} \right)=tan\left( \frac{\pi }{6} \right)\implies \frac{1}{\sqrt{3}} \\------------------\\ thus\implies \int\limits_{0}^{\frac{1 }{\sqrt{3}}}u^2\cdot du$

and surely you can take it from there,
recall, that, since we changed the bounds, with the u(x),
you don't need to change the variable "u", and simply,
get the integral of it, simple enough, and apply those bounds