Question

Solve ${\int }_0^{\pi }\frac{sinx}{sinx+cosx}dx$

Answer 1

${\int }_0^{\pi }\frac{\sin x}{\sin x+\cos x}dx$

The solution to integration of the given type is given by substituting the numerator as

$\sin x=A(\sin x+\cos x)+B\left[\frac{d}{dx}\right(\sin x+\cos x\left)\right]$............(A)

{$Numerator=A\left(Denominator\right)+B\left[\frac{d}{dx}\right(denominator\left)\right]$ ($a,b$ are constants ) }

$\Rightarrow \sin x=A(\sin x+\cos x)+B[\cos x-\sin x]$

$\Rightarrow \sin x=A(\sin x+\cos x)+B(\cos x-\sin x)$

$\Rightarrow \sin x=A\sin x+A\cos x+B\cos x-B\sin x$

$\Rightarrow \sin x=(A-B)\sin x+(A+B)\cos x$

[ taking common terms]

Now comparing both the sides of the equation for sine & cosine values

Therefore $A-B=1$......(1) (since the coefficient of $\sin x$ is $1$ on LHS )

$A+B=0$.......(2) ( No $\cos x$ terms on LHS)

solving (1) & (2) we have

$A-B=1$

$A+B=0$

Add (1) & (2)

$2A=1$

$A=1/2,B=-1/2$

Now substituting $\sin x$ according to equation (A)

${\int }_0^{\pi }1/2\frac{(\sin x+\cos x)-1/2(\cos x-\sin x)dx}{\sin x+\cos x}$

squaring terms

${\int }_0^{\pi }1/2\frac{(\sin x+\cos x)dx}{\sin x+\cos x}-{\int }_0^{\pi }1/2\frac{\left(\cos x-\sin x\right)dx}{\sin x+\cos x}$

$1/2{\int }_0^{\pi }dx-1/2{\int }_0^{\pi }\frac{\cos x-\sin xdx}{\sin x+\cos x}$

$1/2[x{]}_0^{\pi }-1/2[\log |\sin x+\cos x|{]}_0^{\pi }\{\int dx=x\int \frac{f^{\prime }\left(x\right)}{fx}=\log \left(fx\right)\}$

$\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x$

$1/2[\pi -0]-1/2\left[\log \right(\sin \pi +\cos \pi )-\log (\sin 0+\cos 0)$

$1/2\pi -1/2\log \frac{(\sin \pi +\cos \pi )}{\sin 0+\cos 0}$ $[\log x-\log y=\log x/y]$

$\frac{\pi }{2}-1/2\log \left(\frac{0-1}{0+1}\right)$

$\pi /2-1/2(-\log 1)$ $[\log 1=0]$

$\pi /2-1/2\left(0\right)$

$=\pi /2$