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Question: The value of the integral \(\int _ { - \pi / 4 } ^ { \pi / 4 } \sin ^ { - 4 } x d x\) is...

The value of the integral ∫−π/4π/4sin⁡−4xdx\int _ { - \pi / 4 } ^ { \pi / 4 } \sin ^ { - 4 } x d x is

A

3/2

B

–8/3

C

3/8

D

8/3

Answer

–8/3

Explanation

Solution

∫−π/4π/4sin⁡−4xdx=2∫0π/4cos⁡4xsin⁡4xsec⁡4xdx=2∫0π/4sec⁡4xdxtan⁡4x\int _ { - \pi / 4 } ^ { \pi / 4 } \sin ^ { - 4 } x d x = 2 \int _ { 0 } ^ { \pi / 4 } \frac { \cos ^ { 4 } x } { \sin ^ { 4 } x } \sec ^ { 4 } x d x = 2 \int _ { 0 } ^ { \pi / 4 } \frac { \sec ^ { 4 } x d x } { \tan ^ { 4 } x }

Put tan⁡x=t\tan x = t , we get 2∫011+t2t4dt2 \int _ { 0 } ^ { 1 } \frac { 1 + t ^ { 2 } } { t ^ { 4 } } d t

=2[∣−13t3∣01+∣−1t∣01]=−83= 2 \left[ \left| - \frac { 1 } { 3 t ^ { 3 } } \right| _ { 0 } ^ { 1 } + \left| - \frac { 1 } { t } \right| _ { 0 } ^ { 1 } \right] = - \frac { 8 } { 3 }.