First slide

Methods of integration

Question

$\int \frac{x^{2} (x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x is equal to$

Moderate

A

$\frac{- x^{2}}{x \tan x + 1} + 2 \log |x \sin x + \cot x| + C$
B

$\frac{x^{2}}{{(x \tan x + 1)}^{2}} - 2 \log |x \sin x + \cot x| + C$
C

$\frac{x^{2}}{x \tan x + 1} + 2 \log |x \sin x - \cos x| + C$
D

$\frac{x^{2}}{x \tan x + 1} + 2 \log |x \sin x + \cos x| + K$

Solution

$Let I = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x$
Using integrating by parts
$I = x^{2} \int \frac{(x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x - \int \frac{d}{d x} (x^{2}) \int \frac{(x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x$
$x \tan x + 1 = t$
$\begin{aligned} (x \sec^{2} x + \tan x) d x = d t \\ I = x^{2} \int \frac{d t}{t^{2}} - \int [2 x \int \frac{d t}{t^{2}}] d x \end{aligned}$
$\begin{array}{l} = \frac{- x^{2}}{t} - \int 2 x (\frac{- 1}{t}) d x \\ = \frac{- x^{2}}{x \tan x + 1} + \int \frac{2 x}{(x \tan x + 1)} d x \\ = \frac{- x^{2}}{x \tan x + 1} + \int \frac{2 x \cos x}{x \sin x + \cos x} d x \end{array}$
$Put x \sin x + \cos x = y$
$\begin{array}{l} (x \cos x + 1 \sin x - \sin x) \\ = \frac{- x^{2}}{x \tan x + 1} + 2 \int \frac{d t}{t} \\ = \frac{- x^{2}}{x \tan x + 1} + 2 \log | x \sin x + \cos x | + C \end{array}$

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