Question

The value of integral, $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}$
( a ) $\dfrac{\pi }{2}(\sqrt{2}+1)$
( b ) $\pi (\sqrt{2}-1)$
( c ) $2\pi (\sqrt{2}-1)$
( d ) $\pi \sqrt{2}$

Answer 1

To solve this question, we will use property of definite integration such as $\int\limits_{a}^{b}{f(x)dx=F(a)-F(b)}$ and $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$. Also, we will use some trigonometric properties and trigonometric values to solve this question also some algebraic identity such as $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$. Hence, using these we can solve the integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}$

Complete step by step answer:
To solve such a question, we must know some properties of definite integration and indefinite integration..
One of the most important property of definite integration is $\int\limits_{a}^{b}{f(x)dx=F(a)-F(b)}$ ,where F is the integration of f ( x ) and a is lower limit and b is upper limit.
Another important property of definite integral is $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$.
Also, $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n}+C}$ , where C is constant which does not appear I definite integral but does appear in indefinite integral.
Now, we have to evaluate $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}$………( i ),
So, using property $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$. We get
a = $\dfrac{\pi }{4}$ and b = $\dfrac{3\pi }{4}$
so, $a\text{ }+\text{ }b\text{ }\text{ }-x\text{ }=\text{ }\dfrac{\pi }{4}+\dfrac{3\pi }{4}-x$
$a\text{ }+\text{ }b\text{ }\text{ }-x\text{ }=\text{ }\pi -x$
So, $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi -x}{1+\sin (\pi -x)}\cdot dx}$
We know that $\sin (\pi -\theta )=\sin \theta$ ,
So, $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi -x}{1+\sin x}\cdot dx}$
$I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi }{1+\sin x}\cdot dx}-\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}$……. ( ii )
Adding ( i ) and ( ii ), we get
$I+I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi }{1+\sin x}\cdot dx}-\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}+\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{x}{1+\sin x}\cdot dx}$
On solving we get
$2I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi }{1+\sin x}\cdot dx}$
Multiplying numerator and denominator by $1-\sin x$, we get
$2I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\pi }{1+\sin x}\times \dfrac{1-\sin x}{1-\sin x}\cdot dx}$
$I=\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{1-\sin x}{(1+\sin x)(1-\sin x)}dx}$
Using, algebraic identity, $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ ,we get
$I=\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{1-\sin x}{(1-{{\sin }^{2}}x)}dx}$
We know that, $1-{{\sin }^{2}}x={{\cos }^{2}}x$ ,
So, $I=\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{1-\sin x}{{{\cos }^{2}}x}dx}$
Or, $I=\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{1}{{{\cos }^{2}}x}dx}-\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\dfrac{\sin x}{{{\cos }^{2}}x}dx}$
Now, as $\cos x=\dfrac{1}{\sec x}$ , $\tan x=\dfrac{\sin x}{cosx}$
So, $I=\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{{{\sec }^{2}}xdx}-\dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\tan x\sec xdx}$
Now, we know that, $\int{{{\sec }^{2}}x\cdot dx=\operatorname{tanx}}$ and $\int{\sec x\tan x\cdot dx=\sec x}$,
So, $I=\dfrac{\pi }{2}{{\left[ \tan x-\sec x \right]}_{\dfrac{\pi }{4}}}^{\dfrac{3\pi }{4}}$
Putting values of upper limit and lower limit, we get
$I=\dfrac{\pi }{2}\left[ \left( \tan \dfrac{3\pi }{4}-\sec \dfrac{3\pi }{4} \right)-\left( \tan \dfrac{\pi }{4}-\sec \dfrac{\pi }{4} \right) \right]$
As, $\tan \dfrac{\pi }{4}=1,\sec \dfrac{\pi }{4}=\sqrt{2}$
$I=\dfrac{\pi }{2}\left[ \left( \tan \left( \pi -\dfrac{\pi }{4} \right)-\sec \left( \pi -\dfrac{\pi }{4} \right) \right)-\left( 1-\sqrt{2} \right) \right]$
As, $\tan \left( \pi -\dfrac{\pi }{4} \right)=-1$and $\sec \left( \pi -\dfrac{\pi }{4} \right)=-\sqrt{2}$
So,$I=\dfrac{\pi }{2}\left[ \left( -1+\sqrt{2} \right)-\left( 1-\sqrt{2} \right) \right]$
On solving, we get
$I=\dfrac{\pi }{2}\cdot 2(\sqrt{2}-1)$
$I=\pi (\sqrt{2}-1)$

So, the correct answer is “Option b”.

Note: While integrating the definite integral always use the correct formula to evaluate the integration and always try to skip calculation error as it may change the answer of the solution or may make the solution more complex. One must know all the properties and formula of indefinite and definite integrals.