Question

∫ Sec^6 t dt from (-π/4) to (π/4)

Answer 1

We define 

$ǀ_n=∫_{-π/4}^{π/4}sec^6 t dt =[\frac {tan tsec^{n-2} t }{n-1}] ^{π/4}_{-π/4} + \frac {n-2}{n-11}ǀ_{n-2},n≥2$

As the reduction formula for the integral

$؞ ǀ_6=∫sec^6 t dt$

$=[\frac {tan t sec ^4 t}{5}]_{-π/4} ^{π/4} + \frac 45 ǀ_4…(i)$

$ǀ_4=[\frac {tan t sec ^2 t}{3}]^{\pi/4} _{-\pi/4}+ \frac 23 ǀ_2…(ii)$

$ǀ_2= [ tan t]^{π/4} _{-π)/4}…(iii)$

Replacing (iii) in (ii) and replacing (ii) in (i) gives

$ǀ_6=[\frac 15 tan t sec^4 t+\frac 4{15} tan t sec ^2 t+ \frac 8{15} tan t]^{π/4} _{-π)/4}$

$∫_{-π)/4}^{π/4}sec^6 t dt= [\frac15 (1) \sqrt2) ^4 + \frac 4{15} (1) \sqrt2)^2+ \frac 8{15} (-1)] - [\frac 15 (-1) sqrt2^4+ \frac 4{15} (-1)\sqrt 2^2 )+\frac 8{15 }(-1)] =\frac {56}{15}=3.73$