Let I ( x ) = x 2 ( x sec 2 x + tan x ) ( x tan x + 1 ) 2 ? d x . If I ( 0 ) = 0 , then I ( 4 ) is equal to [2023]

Question

$Let I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x . If I (0) = 0, then I (\frac{π}{4}) is equal to$ [2023]

Smart Achievers · Accepted Answer

(4)

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

By using integration by parts

$I (x) = x^{2} \int \frac{(x \sec^{2} x + \tan x) d x}{{(x \tan x + 1)}^{2}} - \int (2 x) \int \frac{(x \sec^{2} x + \tan x) d x}{{(x \tan x + 1)}^{2}}$

$I (x) = x^{2} \frac{(- 1)}{(x \tan x + 1)} + \int \frac{2 x}{x \tan x + 1} d x$

$I (x) = x^{2} (\frac{- 1}{x \tan x + 1}) + 2 \int \frac{x \cos x}{x \sin x + \cos x} d x$

$I (x) = x^{2} (\frac{- 1}{x \tan x + 1}) + 2 \ln$ $| x \sin x + \cos x |$ $+ C$

$As, I (0) = 0$

$\Rightarrow 0 = 0 + 2 \ln | 1 | + C$ $∴$ $C = 0$

$∴$ $I (\frac{π}{4}) = \frac{- π^{2}}{16} (\frac{1}{\frac{π}{4} + 1}) + 2 \ln$ $| \frac{π}{4} \sin (\frac{π}{4}) + \cos \frac{π}{4} |$

$= 2 \ln (\frac{(π + 4)}{4 \sqrt{2}}) - \frac{π^{2}}{4 (π + 4)} = \ln [\frac{{(π + 4)}^{2}}{32}] - \frac{π^{2}}{4 (π + 4)}$

Solution

Q.
$Let I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x . If I (0) = 0, then I (\frac{π}{4}) is equal to$ [2023]

1 $\log_{e} \frac{{(π + 4)}^{2}}{16} + \frac{π^{2}}{4 (π + 4)}$

2 $\log_{e} \frac{{(π + 4)}^{2}}{32} + \frac{π^{2}}{4 (π + 4)}$

3 $\log_{e} \frac{{(π + 4)}^{2}}{16} - \frac{π^{2}}{4 (π + 4)}$

4 $\log_{e} \frac{{(π + 4)}^{2}}{32} - \frac{π^{2}}{4 (π + 4)}$

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Solution

Q. Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2 dx. If I(0)=0, then I(π4) is equal to [2023]

1 loge(π+4)216+π24(π+4)

2 loge(π+4)232+π24(π+4)

3 loge(π+4)216-π24(π+4)