## Calculus, 10th Edition (Anton)

(1/4)$sec^{3}$x·tanx+(3/8)ln|tanx + secx|-(5/8)tanx·secx +C
$\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$$sec^{5}x dx + \intsecx dx -2 \int$$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$$sec^{3}x dx=\intsecx (sec^{2}x) dx Let u=secx & dv= (sec^{2}x) dx, so v=tanx \int$$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$$secx dx ==> \int$$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