Solveeit Logo
Login

Question

Question: \(\int _ { 0 } ^ { \pi / 4 } \frac { \sec ^ { 2 } x } { ( 1 + \tan x ) ( 2 + \tan x ) } d x =\)...

∫0π/4sec⁡2x(1+tan⁡x)(2+tan⁡x)dx=\int _ { 0 } ^ { \pi / 4 } \frac { \sec ^ { 2 } x } { ( 1 + \tan x ) ( 2 + \tan x ) } d x =

A

log⁡e(23)\log _ { e } \left( \frac { 2 } { 3 } \right)

B

log⁡e3\log _ { e } 3

C

12log⁡e(43)\frac { 1 } { 2 } \log _ { e } \left( \frac { 4 } { 3 } \right)

D

log⁡e(43)\log _ { e } \left( \frac { 4 } { 3 } \right)

Answer

log⁡e(43)\log _ { e } \left( \frac { 4 } { 3 } \right)

Explanation

Solution

Put 1+tan⁡x=t⇒sec⁡2xdx=dt1 + \tan x = t \Rightarrow \sec ^ { 2 } x d x = d t

∴∫0π/4sec⁡2x(1+tan⁡x)(2+tan⁡x)dx\therefore \int _ { 0 } ^ { \pi / 4 } \frac { \sec ^ { 2 } x } { ( 1 + \tan x ) ( 2 + \tan x ) } d x

=∫12dtt(1+t)=∫12dtt−∫12dt1+t=[log⁡t−log⁡(1+t)]12= \int _ { 1 } ^ { 2 } \frac { d t } { t ( 1 + t ) } = \int _ { 1 } ^ { 2 } \frac { d t } { t } - \int _ { 1 } ^ { 2 } \frac { d t } { 1 + t } = [ \log t - \log ( 1 + t ) ] _ { 1 } ^ { 2 }

=log⁡e2−log⁡e3+log⁡e2=log⁡e43= \log _ { e } 2 - \log _ { e } 3 + \log _ { e } 2 = \log _ { e } \frac { 4 } { 3 }.