Question

integration dx/cos^2x+cot^2x
​

Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x+\cot^2x}$

Since $\cot x=\dfrac{\cos x}{\sin x},$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x+\dfrac{\cos^2x}{\sin^2x}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x\left(1+\dfrac{1}{\sin^2x}\right)}$

$\displaystyle\longrightarrow I=\int\dfrac{\sec^2x\ dx}{\left(1+\dfrac{1}{\sin^2x}\right)}$

$\displaystyle\longrightarrow I=\int\dfrac{\sin^2x}{1+\sin^2x}\cdot\sec^2x\ dx$

Dividing both numerator and denominator of $\dfrac{\sin^2x}{1+\sin^2x}$ by $\cos^2x,$

$\displaystyle\longrightarrow I=\int\dfrac{\left(\dfrac{\sin^2x}{\cos^2x}\right)}{\left(\dfrac{1+\sin^2x}{\cos^2x}\right)}\cdot\sec^2x\ dx$

$\displaystyle\longrightarrow I=\int\dfrac{\tan^2x}{\sec^2x+\tan^2x}\cdot\sec^2x\ dx$

Since $\sec^2x=1+\tan^2x,$

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

Substitute,

$\longrightarrow u=\tan x$

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

Then,

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

Multiplying and dividing by 2,

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

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

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

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

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2}\int\dfrac{1}{1^2+\left(u\sqrt2\right)^2}\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2}\cdot\dfrac{1}{\sqrt2}\,\tan^{-1}\left(u\sqrt2\right)+C$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2\sqrt2}\,\tan^{-1}\left(u\sqrt2\right)+C$

Undoing substitution $u=\tan x,$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}\tan x-\dfrac{1}{2\sqrt2}\tan^{-1}\left(\sqrt2\,\tan x\right)+C}}$

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

We're asked to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x+\cot^2x}$

Since $\cot x=\dfrac{\cos x}{\sin x},$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x+\dfrac{\cos^2x}{\sin^2x}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\cos^2x\left(1+\dfrac{1}{\sin^2x}\right)}$

$\displaystyle\longrightarrow I=\int\dfrac{\sec^2x\ dx}{\left(1+\dfrac{1}{\sin^2x}\right)}$

$\displaystyle\longrightarrow I=\int\dfrac{\sin^2x}{1+\sin^2x}\cdot\sec^2x\ dx$

Dividing both numerator and denominator of $\dfrac{\sin^2x}{1+\sin^2x}$ by $\cos^2x,$

$\displaystyle\longrightarrow I=\int\dfrac{\left(\dfrac{\sin^2x}{\cos^2x}\right)}{\left(\dfrac{1+\sin^2x}{\cos^2x}\right)}\cdot\sec^2x\ dx$

$\displaystyle\longrightarrow I=\int\dfrac{\tan^2x}{\sec^2x+\tan^2x}\cdot\sec^2x\ dx$

Since $\sec^2x=1+\tan^2x,$

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

Substitute,

$\longrightarrow u=\tan x$

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

Then,

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

Multiplying and dividing by 2,

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

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

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

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

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2}\int\dfrac{1}{1^2+\left(u\sqrt2\right)^2}\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2}\cdot\dfrac{1}{\sqrt2}\,\tan^{-1}\left(u\sqrt2\right)+C$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\,u-\dfrac{1}{2\sqrt2}\,\tan^{-1}\left(u\sqrt2\right)+C$

Undoing substitution $u=\tan x,$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}\tan x-\dfrac{1}{2\sqrt2}\tan^{-1}\left(\sqrt2\,\tan x\right)+C}}$