What is the integral of $\int {tan}^{6} (x) {sec}^{6} (x)$ $int tan^6(x)sec^6(x)$ ?

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What is the integral of $\int {tan}^{6} (x) {sec}^{6} (x)$ $int tan^6(x)sec^6(x)$ ?

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Answer 1 · 2016-03-25T20:39:45

$\frac{{tan}^{11} (x)}{11} + \frac{2 {tan}^{9} (x)}{9} + \frac{{tan}^{7} (x)}{7} + C$ $tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C$

Explanation:

We will want to approach this so as to leave a ${sec}^{2} (x)$ $sec^2(x)$ term hanging around in the integral. $($ $($ This will act as $d u$ $du$ if we set $u = tan (x))$ $u=tan(x))$ . If we convert the remaining ${sec}^{4} (x)$ $sec^4(x)$ into functions of $tan (x)$ $tan(x)$ through Pythagorean identities, then we can use $u$ $u$ substitution.

$\int {tan}^{6} (x) {sec}^{6} (x) d x$ $inttan^6(x)sec^6(x)dx$

$= \int {tan}^{6} (x) {sec}^{4} (x) {sec}^{2} (x) d x$ $=inttan^6(x)sec^4(x)sec^2(x)dx$

$= \int {tan}^{6} (x) {({sec}^{2} (x))}^{2} {sec}^{2} (x) d x$ $=inttan^6(x)(sec^2(x))^2sec^2(x)dx$

$= \int {tan}^{6} (x) {(1 + {tan}^{2} (x))}^{2} {sec}^{2} (x) d x$ $=inttan^6(x)(1+tan^2(x))^2sec^2(x)dx$

$= \int {tan}^{6} (x) (1 + 2 {tan}^{2} (x) + {tan}^{4} (x)) {sec}^{2} (x) d x$ $=inttan^6(x)(1+2tan^2(x)+tan^4(x))sec^2(x)dx$

$= \int ({tan}^{10} (x) + 2 {tan}^{8} (x) + {tan}^{6} (x)) {sec}^{2} (x) d x$ $=int(tan^10(x)+2tan^8(x)+tan^6(x))sec^2(x)dx$

Now, let $u = tan x$ $u=tanx$ which implies $d u = {sec}^{2} (x) d x$ $du=sec^2(x)dx$ .

Substituting, we see that

$= \int (u^{10} + 2 u^{8} + u^{6}) d u$ $=int(u^10+2u^8+u^6)du$

Integrating term by term, this gives

$= \frac{u^{11}}{11} + \frac{2 u^{9}}{9} + \frac{u^{7}}{7} + C$ $=u^11/11+(2u^9)/9+u^7/7+C$

Since $u = tan (x)$ $u=tan(x)$ :

$= \frac{{tan}^{11} (x)}{11} + \frac{2 {tan}^{9} (x)}{9} + \frac{{tan}^{7} (x)}{7} + C$ $=tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C$

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What is the integral of $\int {tan}^{6} (x) {sec}^{6} (x)$ $int tan^6(x)sec^6(x)$ ?

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