Question

If $A={\int }_0^{\pi }\frac{sinx}{sinx+cosx}dx$

$B={\int }_0^{\pi }\frac{sinx}{sinx-cosx}dx$

• A. value of A is equal to B
• B. A and B both are rational number
• C. A and B both are irrational number
• D. A is period of |sin x|

Answer 1

Answer: (A), (C)

The correct options are
A

value of A is equal to B

C

A and B both are irrational number

$A={\int }_0^{\pi }\frac{sin(\pi -x)}{sin(\pi -x)+cos(\pi -x)}dx$

$A={\int }_0^{\pi }\frac{sinx}{sinx-cosx}dx=B$

A = B

Now $A+B={\int }_0^{\pi }\frac{sinx(sinx-cosx)+sinx(sinx+cosx)}{(sinx-cosx)(sinx+cosx)}$

$A+B={\int }_0^{\pi }\frac{2si{n}^{2}x}{(si{n}^{2}x-co{s}^{2}x)}dx$

$A+B={\int }_0^{\pi }\frac{1-cos2x}{-cos2x}dx={\int }_0^{\pi }(1-sec2x)dx$

$A+B={[x-\frac{1}{2}log|sec2x+tan2x|]}_0^{\pi }$

$A+B=\pi -\frac{1}{2}log|sec2\pi +tan2\pi |+\frac{1}{2}log|sec0+tan0|$

$A+B=\pi \Rightarrow A=B=\frac{\pi }{2}$