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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 73 of 160

Marks: +1, -0

If

f(x)=\int\;\frac{2-3\sin^{2}x}{1+\cos 2x}dx

f\left(\frac{\pi}{4}\right)=1

f(0)=

$\frac{3}{8}(4-\pi)$
$3-\frac{\pi}{4}$
0
1

Solution:

(a)

f(x)=\int\;\frac{2-3\sin^{2}x}{1+\cos 2x}dx

f(x)=\int\;\frac{2-3\sin^{2}x}{2\cos^{2}x}dx

=\int[\sec^{2}x-\frac{3}{2}\tan^{2}x]dx

=\int[\sec^{2}x-\frac{3}{2}(\sec^{2}x-1)]dx

=\int[\sec^{2}x-\frac{3}{2}\sec^{2}x+\frac{3}{2}]dx

=\int[-\frac{1}{2}\sec^{2}x+\frac{3}{2}]dx

f(x)=-\frac{1}{2}\tan x+\frac{3}{2}x+c

......(i)

f\left(\frac{\pi}{4}\right)=-\frac{1}{2}\tan\frac{\pi}{4}+\frac{3}{2}\cdot\frac{\pi}{4}+c

1=-\frac{1}{2}+\frac{3\pi}{8}+c

\Rightarrow\;\;c=\frac{3}{2}-\frac{3\pi}{8}

\Rightarrow\;\;c=\frac{3}{8}(4-\pi)

From Eq. (i), we have

f(x)=-\frac{1}{2}\tan x+\frac{3}{2}x+\frac{3}{8}(4-\pi)

f(0)=0+0+\frac{3}{8}(4-\pi)=\frac{3}{8}(4-\pi)

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