QJEEM18Mabd3b: JEE Main 2018 Mathematics - Class 12 JEEM18Mabd3b Solution: The value of integral is :

Question

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The value of integral

\int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}

is :

Accepted Answer

Let

I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}

also let K =

{x \over {1 + \sin x}}

Multiplying numerator and denominator by
(1

-

sin x) we get;

K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}

= x(1

-

sin x) sec²x

= x sec²x

-

x sin x sec²x = x sec²x

-

x tan x sec x

Now,

I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} }

\Rightarrow

I =

\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}

-

\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}}

\Rightarrow

I =

\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}

-

\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}

- \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}

\Rightarrow

I =

- {{3\pi } \over 4} - {\pi \over 4}

+ {{3\pi } \over 4}\left( {\sqrt 2 } \right)

+ {\pi \over 4}\left( {\sqrt 2 } \right)

- \log \left( {\sqrt 2 + 1} \right)

+ \log \left( {\sqrt 2 + 1} \right)

\Rightarrow

I = -

\pi

+

{\sqrt 2 \pi }

\Rightarrow

I =

\pi \left( {\sqrt 2 - 1} \right)

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