Question

Area bounded by the curves y = tanx and y = tan²x in between x ∈$\left( - \frac { \pi } { 3 } , \frac { \pi } { 3 } \right)$is equal to

• A. $\frac { 1 } { 2 }$(π + ln 2 – 2) sq. units
• B. $\frac { 1 } { 3 }$(π + ln(2$\sqrt { 2 }$- 3) sq. units
• C. $\frac { 1 } { 4 }$(π + ln4 – 4) sq. units
• D. $\frac { 1 } { 2 }$(π + ln4 – 2) sq. units

Answer 1

Answer: (C)

$\frac { 1 } { 4 }$(π + ln4 – 4) sq. units

Answer 2

The given curves intersect at $x = \frac { \pi } { 4 }$, in between x e$\left( - \frac { \pi } { 3 } , \frac { \pi } { 3 } \right)$

Required area $\Delta = \int _ { 0 } ^ { \pi / 4 } \left( \tan x - \tan ^ { 2 } x \right) d x$

= ln sec x $\left. \right| _ { 0 } ^ { \pi / 4 } - \left. ( \tan x - x ) \right| _ { 0 } ^ { \pi / 4 }$

= $\ln \sqrt { 2 } - \left( 1 - \frac { \pi } { 4 } \right) = \left( \frac { \pi } { 4 } + \ln \sqrt { 2 } - 1 \right)$ sq. units.