What is $\int {tan}^{2} (x) \cdot {sec}^{3} (x) d x$ $int tan^2(x) * sec^3(x) dx$ ?

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What is $\int {tan}^{2} (x) \cdot {sec}^{3} (x) d x$ $int tan^2(x) * sec^3(x) dx$ ?

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Answer 1 · 2016-01-07T01:24:08

$\int {tan}^{2} (x) {sec}^{3} (x) d x = \frac{2 {sec}^{3} (x) tan (x) - sec (x) tan (x) - ln | sec (x) + tan (x) |}{8}$ $inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/8$

Explanation:

$I = \int {tan}^{2} (x) \cdot {sec}^{3} (x) d x$ $I = inttan^2(x)*sec^3(x)dx$
$I = \int ({sec}^{2} (x) - 1) \cdot {sec}^{3} (x) d x$ $I = int(sec^2(x)-1)*sec^3(x)dx$
$I = \int {sec}^{5} (x) d x - \int {sec}^{3} (x) d x$ $I = intsec^5(x)dx - intsec^3(x)dx$

For the second integral, you need integration by parts, say
$u = sec (x)$ $u = sec(x)$ , so $d u = sec (x) tan (x)$ $du = sec(x)tan(x)$ and
$d v = {sec}^{2} (x)$ $dv = sec^2(x)$ so $v = tan (x)$ $v = tan(x)$

$\int {sec}^{3} (x) d x = sec (x) tan (x) - \int sec (x) {tan}^{2} (x) d x$ $intsec^3(x)dx = sec(x)tan(x) - intsec(x)tan^2(x)dx$
$\int {sec}^{3} (x) d x = sec (x) tan (x) - \int sec (x) ({sec}^{2} (x) - 1) d x$ $intsec^3(x)dx = sec(x)tan(x) - intsec(x)(sec^2(x)-1)dx$
$\int {sec}^{3} (x) d x = sec (x) tan (x) - \int ({sec}^{3} (x) - sec (x)) d x$ $intsec^3(x)dx = sec(x)tan(x) - int(sec^3(x)-sec(x))dx$
$\int {sec}^{3} (x) d x = sec (x) tan (x) - \int {sec}^{3} (x) d x - \int sec (x) d x$ $intsec^3(x)dx = sec(x)tan(x) - intsec^3(x)dx -intsec(x)dx$
$2 \int {sec}^{3} (x) d x = sec (x) tan (x) - \int sec (x) d x$ $2intsec^3(x)dx = sec(x)tan(x) - intsec(x)dx$

The last integral is a tabled one, so

$\int {sec}^{3} (x) d x = \frac{sec (x) tan (x) + ln | sec (x) + tan (x) |}{2} + c$ $intsec^3(x)dx = (sec(x)tan(x)+ln|sec(x)+tan(x)|)/2 + c$

For $\int {sec}^{5} d x$ $intsec^5dx$ , using integration by parts
$u = {sec}^{3} (x)$ $u = sec^3(x)$ so $d u = 3 tan (x) {sec}^{3} (x)$ $du = 3tan(x)sec^3(x)$
$d v = {sec}^{2} (x)$ $dv = sec^2(x)$ so $v = tan (x)$ $v = tan(x)$

$\int {sec}^{5} d x = {sec}^{3} (x) tan (x) - \int tan (x) \cdot 3 tan (x) {sec}^{3} (x) d x$ $intsec^5dx = sec^3(x)tan(x) - inttan(x)*3tan(x)sec^3(x)dx$
$\int {sec}^{5} d x = {sec}^{3} (x) tan (x) - 3 \int {tan}^{2} (x) {sec}^{3} (x) d x$ $intsec^5dx = sec^3(x)tan(x) - 3inttan^2(x)sec^3(x)dx$

But we originally had that

$\int {tan}^{2} (x) {sec}^{3} (x) d x = \int {sec}^{5} (x) d x - \int {sec}^{3} (x) d x$ $inttan^2(x)sec^3(x)dx = intsec^5(x)dx - intsec^3(x)dx$

So substituting that we have

$\int {tan}^{2} (x) {sec}^{3} (x) d x = {sec}^{3} (x) tan (x) - 3 \int {tan}^{2} (x) {sec}^{3} (x) d x - \frac{sec (x) tan (x) + ln | sec (x) + tan (x) |}{2}$ $inttan^2(x)sec^3(x)dx = sec^3(x)tan(x) - 3inttan^2(x)sec^3(x)dx - (sec(x)tan(x)+ln|sec(x)+tan(x)|)/2$
$4 \int {tan}^{2} (x) {sec}^{3} (x) d x = \frac{2 {sec}^{3} (x) tan (x) - sec (x) tan (x) - ln | sec (x) + tan (x) |}{2}$ $4inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/2$
$\int {tan}^{2} (x) {sec}^{3} (x) d x = \frac{2 {sec}^{3} (x) tan (x) - sec (x) tan (x) - ln | sec (x) + tan (x) |}{8}$ $inttan^2(x)sec^3(x)dx = (2sec^3(x)tan(x) - sec(x)tan(x)-ln|sec(x)+tan(x)|)/8$

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What is $\int {tan}^{2} (x) \cdot {sec}^{3} (x) d x$ $int tan^2(x) * sec^3(x) dx$ ?

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What is int tan^2(x) * sec^3(x) dx ∫tan2(x)⋅sec3(x)dxint tan^2(x) * sec^3(x) dx ?

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What is $\int {tan}^{2} (x) \cdot {sec}^{3} (x) d x$ $int tan^2(x) * sec^3(x) dx$ ?