Correct Answer is : 12log⁡∣tan⁡(x2+π8)∣+C{1}{{2}}≤ft|≤ft(x/2+π/8)|+C21logtan(2x+8π)+C

{1}{{2}}≤ft|≤ft(x/2+π/8)|+C

{1}{{2}}≤ft|≤ft(x/2+π/4)|+C

{1}{{2}}≤ft|≤ft(x/4+π/2)|+C

{1}{{2}}≤ft|≤ft(x/8+π/2)|+C

Correct Answer is : 12log⁡∣tan⁡(x2+π8)∣+C{1}{{2}}≤ft|≤ft(x/2+π/8)|+C21logtan(2x+8π)+C

Test Index

TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int\frac{dx}{\sin x+\cos x}

$\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{\pi}{8}\right)\right|+C$
$\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{\pi}{4}\right)\right|+C$
$\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{4}+\frac{\pi}{2}\right)\right|+C$
$\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{8}+\frac{\pi}{2}\right)\right|+C$

Solution: 👈: Video Solution

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

\left[\begin{array}{l} \sin 2x = \frac{2\tan x}{1+\tan^2 x} \\ \because \quad \cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x} \end{array}\right]

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

=\int\;\frac{\sec^2\frac{x}{2}dx}{2\tan\frac{x}{2}+1-\tan^2\frac{x}{2}}

Put

\tan\frac{x}{2}=t \Rightarrow \sec^2\frac{x}{2}dx=2dt

=2\int\frac{dt}{2t+1-t^2}=-2\int\frac{dt}{t^2-2t-1}

=-2\int\frac{dt}{(t-1)^2-(\sqrt{2})^2}

=2\int\frac{dt}{(\sqrt{2})^2-(t-1)^2}

=2\cdot\frac{1}{2\sqrt{2}}\log\left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|

=\frac{1}{\sqrt{2}}\log\left|\frac{\sqrt{2}+\tan\frac{x}{2}-1}{\sqrt{2}-\tan\frac{x}{2}+1}\right|+C

=\frac{1}{\sqrt{2}}\log\left|\frac{\tan\frac{x}{2}+(\sqrt{2}-1)}{1-\tan\frac{x}{2}+\sqrt{2}}\right|+C

\tan\frac{\pi}{8} = \sqrt{2}-1, \cot\frac{\pi}{8} = \sqrt{2}+1

=\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{\pi}{8}\right)\right|+C

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