The value of integral _/4^3/4 x1+ x,dx is

Question

The value of integral $\displaystyle\int_{\pi/4}^{3\pi/4} \dfrac{x}{1+\sin x}\,dx$ 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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