Answer
(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx + secx|-(5/8)tanx·secx +C
Work Step by Step
$\int$$(tan^{4}$x) (secx)dx
Step 1: Use the identity: $tan^{2}$x =$sec^{2}$x -1
$\int$$(tan^{4}$x) (secx)dx=$\int$$(tan^{2}x)^{2}$secx dx=$\int$$(sec^{2}x-1)^{2}$ secx dx
Expand $(sec^{2}x-1)^{2}$ secx :
the expression= $\int$$sec^{5}$x dx + $\int$secx dx -2 $\int$$sec^{3}$x dx
= $\int$$sec^{5}$x dx + ln|secx + tanx|-2 $\int$$sec^{3}$x dx (1)
Step 2: In expression (1), $\int$$sec^{3}$x dx=$\int$secx ($sec^{2}$x) dx
Let u=secx & dv= ($sec^{2}$x) dx, so v=tanx
$\int$$sec^{3}$x dx=secx(tanx)-$\int$$(tan^{2}$x) (secx)dx
As $tan^{2}$x =$sec^{2}$x -1, the expression =secx(tanx)-$\int$$(sec^{2}$x-1) (secx)dx
Split the integral: the expression $\int$$sec^{3}$x dx= secx(tanx)-$\int$$sec^{3}$xdx+$\int$$sec$x dx
==> $\int$$sec^{3}$x dx =secx(tanx)-$\int$$sec^{3}$xdx+ln|tanx + secx| (2)
Solve for $\int$$sec^{3}$x dx in equation (2): $\int$$sec^{3}$x dx=(1/2)(ln|tanx + secx|+ tanx·secx)
Step 3: In expression (1), $\int$$sec^{5}$x dx=$\int$($sec^{3}$x) ($sec^{2}$x) dx
Let u=$sec^{3}$x & dv= ($sec^{2}$x) dx, so v=tanx
$\int$$sec^{5}$x dx=$sec^{3}$x(tanx)-3$\int$$(tan^{2}$x) ($sec^{3}$x)dx
Use the identity $tan^{2}$x =$sec^{2}$x -1,
The expression = $sec^{3}$x(tanx)-3$\int$$(sec^{2}$x-1)($sec^{3}$x)dx
And split the integral: $\int$$sec^{5}$x dx =$sec^{3}$x(tanx)-3$\int$$sec^{5}$x dx+3$\int$$sec^{3}$x dx
From step (2) we get $\int$$sec^{3}$x dx=(1/2)(ln|tanx + secx|+ tanx·secx)
So $\int$$sec^{5}$x dx =$sec^{3}$x(tanx)-3$\int$$sec^{5}$x dx+(3/2)(ln|tanx + secx|+ tanx·secx) (3)
Solve for $\int$$sec^{5}$x dx in equation (3):
$\int$$sec^{5}$x dx=(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx+secx|+(3/8)tanx·secx
Thus, combining expressions:
(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx + secx|-(5/8)tanx·secx +C
