Question

If f(x) = $\frac { x ^ { 9 } - 3 x ^ { 5 } + 7 x ^ { 3 } - x + 1 } { \cos ^ { 2 } x }$ then $\int _ { - \pi / 4 } ^ { \pi / 4 } \mathrm { f } ( \mathrm { x } ) \mathrm { dx }$ equal to –

• A. 0
• B. p/2
• C. 2
• D. 1

Answer 1

Answer: (C)

2

Answer 2

Hence f(x) = $\frac { x ^ { 9 } - 3 x ^ { 5 } + 7 x ^ { 3 } - 7 x } { \cos ^ { 2 } x }$ + sec²x

Ž (x⁹ – 3x⁵ + 7x³ – x) +

Ž odd +

Ž $2 [ \tan \mathrm { x } ] _ { 0 } ^ { \pi / 4 }$ = 2$\left[ \tan \frac { \pi } { 4 } - \tan 0 \right]$ = 2