Question

integrate it???.....​

Answer 1

Answer:

sec^2

Step-by-step explanation:

this general identity of trignometric

Answer 2

Given to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^{\pi}\dfrac{1}{1+2\tan x}\ dx$

$\displaystyle\longrightarrow I=\int\limits_0^{\pi}\dfrac{\sec^2x\ dx}{(1+2\tan x)(1+\tan^2x)}\quad\quad\dots(1)$

Substitute,

$\longrightarrow u=\tan x$

$\longrightarrow du=\sec^2\ dx$

Then (1) becomes, (excluding the limits)

$\displaystyle\longrightarrow I=\int\dfrac{1}{(1+2u)(1+u^2)}\ du\quad\quad\dots(2)$

Let,

$\longrightarrow\dfrac{1}{(1+2u)(1+u^2)}=\dfrac{2A}{1+2u}+\dfrac{2Bu+C}{1+u^2}$

$\longrightarrow 2A(1+u^2)+(2Bu+C)(1+2u)=1$

$\longrightarrow2(A+2B)u^2+2(B+C)u+(2A+C)=1$

Equating corresponding coefficients,

$\longrightarrow A+2B=0$

$\longrightarrow B+C=0$

$\longrightarrow 2A+C=1$

Solving them we get,

$\longrightarrow A=\dfrac{2}{5}$

$\longrightarrow B=-\dfrac{1}{5}$

$\longrightarrow C=\dfrac{1}{5}$

Then (4) becomes,

$\displaystyle\longrightarrow I=\int\left(\dfrac{2\cdot\frac{2}{5}}{1+2u}+\dfrac{2\cdot-\frac{1}{5}u+\frac{1}{5}}{1+u^2}\right)\ du$

$\displaystyle\longrightarrow I=\dfrac{2}{5}\int\dfrac{2\ du}{1+2u}-\dfrac{1}{5}\int\dfrac{2u\ du}{1+u^2}+\dfrac{1}{5}\int\dfrac{1}{1+u^2}\ du$

$\displaystyle\longrightarrow I=\dfrac{2}{5}\log|1+2u|-\dfrac{1}{5}\log\left(1+u^2\right)+\dfrac{1}{5}\tan^{-1}u$

$\displaystyle\longrightarrow I=\dfrac{1}{5}\left[\log\left(\dfrac{(1+2u)^2}{1+u^2}\right)+\tan^{-1}u\right]$

Undoing $u=\tan x,$ (including the limits)

$\displaystyle\longrightarrow I=\dfrac{1}{5}\left[\log\left(\dfrac{(1+2\tan x)^2}{1+\tan^2x}\right)+x\right]_0^{\pi}$

$\displaystyle\longrightarrow I=\dfrac{1}{5}\left[\log\left(\dfrac{(1+0)^2}{1+0^2}\right)+\pi-\log\left(\dfrac{(1+0)^2}{1+0^2}\right)-0\right]$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{\pi}{5}}}$