Question

Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx

Answer 1

If we use the substitution $u = \tan x$, then $du = \sec^2 {x}\ dx$. If you try substituting just $u$ and $du$ into the integrand, though, you'll notice that there's a $\sec^2x$ left over that we have to deal with.

To get rid of this problem, use the identity $\tan^2 x + 1 = \sec^2 x$ and substitute in the left side of the identity for the extra $\sec^2x$, as shown:

$\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx$
$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$

From there, we can substitute in $u$ and $du$, and then evaluate:

$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du$
$= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}$
$= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})$
$= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}$