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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 72 of 160

Marks: +1, -0

\int \frac{x^2 (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx = A \log(|x \sin x + \cos x|) + B \frac{f(x)}{x \tan x + 1} + C

f(A+B)=

1
0
-1
2

Solution:

Let

I = \int x^{2} \cdot \frac{x \sec^{2} x + \tan x}{(x \tan x + 1)^{2}} dx

= x^{2} \left(-\frac{1}{x \tan x + 1}\right) - \int 2x \left(-\frac{1}{x \tan x + 1}\right) dx

[

since

\int \frac{x \sec^{2} x + \tan x}{(x \tan x + 1)^{2}} = \int \frac{dt}{t^{2}} = -\frac{1}{t} = -\frac{1}{x \tan x + 1} ]

= - \frac{x^{2}}{x \tan x + 1} + \int \frac{2x \cos x}{x \sin x + \cos x} dx

[putting

x \sin x + \cos x = u

\Rightarrow (x \cos x + \sin x - \sin x) dx = du ]

= \frac{-x^{2}}{x \tan x + 1} + 2 \int \frac{du}{u}

= \frac{-x^{2}}{x \tan x + 1} + 2 \log |u| + C

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

\therefore A = 2

and

B = -1

and

f(x) = x^{2}

\therefore f(A+B) = f(2-1) = f(1) = (1)^{2} = 1

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