Question

limit 0 to π/2 ∫ tan^8x + tan^10x dx

Answer 1

Answer:

limit 0 to π/2 ∫ tan^8x + tan^10x dx =∞

Step-by-step explanation:

To integrate

$\int \: ( {tan}^{8} x + {tan}^{10} x)dx \\ \\ \int \: {tan}^{8} x(1 + {tan}^{2} x)dx \\ \\ \because \: 1 + {tan}^{2} x = {sec}^{2} x \\ \\ \int \: {tan}^{8} x \: {sec}^{2} x \: dx \\ \\ \: let \: \: tan \: x = t \\ \\ {sec}^{2} x \: dx = dt \\ \\ \int \: {t}^{8} \: dt = \frac{ {t}^{9} }{9} + c \\ \\ redo \: substitution \\ \\ \int \: ( {tan}^{8} x + {tan}^{10} x)dx = \frac{ {tan}^{9} x}{9} + c$

Now to integrate the function from 0 to π/2,place upper and lower limit in the RHS

$\int\limit_{0}^{\frac{\pi}{2}}( {tan}^{8} x + {tan}^{10} x) \, dx =\Big[\frac{ {tan}^{9} x}{9}\Big]\limit_{0}^{\frac{\pi}{2}}\\\\ =\frac{ {tan}^{9}\frac{\pi}{2} }{9} -\frac{ {tan}^{9}0}{9}=\infty$

Hope it helps you.

Answer 2

Answer:

∞

Step-by-step explanation:

To integrate  

$\int \: ( {tan}^{8} x +  {tan}^{10} x)dx$  

$\int \:  {tan}^{8} x(1 +  {tan}^{2} x)dx$

$\because \: 1 + {tan}^{2} x =  {sec}^{2} x$  

$\int \:  {tan}^{8} x \:  {sec}^{2} x \: dx$

$\: let \:  \: tan \: x = t$  

${sec}^{2} x \: dx = dt$

$\int \:  {t}^{8}  \: dt  = \frac{ {t}^{9} }{9}  + c$

$redo \: substitution$

$\int \: ( {tan}^{8} x +  {tan}^{10} x)dx = \frac{ {tan}^{9} x}{9}  + c$

Now to integrate the function from 0 to π/2,place upper and lower limit in the RHS

$\int\limit_{0}^{\frac{\pi}{2}}( {tan}^{8} x +  {tan}^{10} x) \, dx =\Big[\frac{ {tan}^{9} x}{9}\Big]\limit_{0}^{\frac{\pi}{2}$

$=\frac{ {tan}^{9}\frac{\pi}{2} }{9} -\frac{ {tan}^{9}0}{9}=\infty$