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∫ sinx/(1+cosx+sinx) dx, 0 to π/2 | JEE

JEE Maths question with a full step-by-step solution.

Question
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Aπ4\dfrac{\pi}{4}
Bπ4+log⁡2\dfrac{\pi}{4}+\log\sqrt{2}
Cπ4−log⁡2\dfrac{\pi}{4}-\log\sqrt{2}correct
Dπ4−log⁡2\dfrac{\pi}{4}-\log 2
Solution
Solution: Step 1: Reduce the denominator with half-angle substitutions. Using 1+cos⁡x=2cos⁡2x21+\cos x=2\cos^{2}\dfrac{x}{2} and sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2},
1+cos⁡x+sin⁡x=2cos⁡x2(cos⁡x2+sin⁡x2),sin⁡x=2sin⁡x2cos⁡x2.1+\cos x+\sin x=2\cos\dfrac{x}{2}\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right),\qquad \sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2}.
 Hence
I=∫0π/22sin⁡x2cos⁡x22cos⁡x2(sin⁡x2+cos⁡x2) dx=∫0π/2sin⁡x2sin⁡x2+cos⁡x2 dx.I=\int_{0}^{\pi/2}\dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos\frac{x}{2}\big(\sin\frac{x}{2}+\cos\frac{x}{2}\big)}\,dx=\int_{0}^{\pi/2}\dfrac{\sin\frac{x}{2}}{\sin\frac{x}{2}+\cos\frac{x}{2}}\,dx.
 Step 2: Put x2=u\dfrac{x}{2}=u so dx=2 dudx=2\,du, with u:0→π4u:0\to\dfrac{\pi}{4}:
I=2∫0π/4sin⁡usin⁡u+cos⁡u du.I=2\int_{0}^{\pi/4}\dfrac{\sin u}{\sin u+\cos u}\,du.
 Step 3: Resolve sin⁡u\sin u in terms of the denominator and its derivative. Write sin⁡u=A(sin⁡u+cos⁡u)+B(cos⁡u−sin⁡u)\sin u=A(\sin u+\cos u)+B(\cos u-\sin u). Matching coefficients gives A−B=1A-B=1 and A+B=0A+B=0, so A=12, B=−12A=\dfrac12,\ B=-\dfrac12. Then
sin⁡usin⁡u+cos⁡u=12−12⋅cos⁡u−sin⁡usin⁡u+cos⁡u.\dfrac{\sin u}{\sin u+\cos u}=\dfrac12-\dfrac12\cdot\dfrac{\cos u-\sin u}{\sin u+\cos u}.
 Step 4: Integrate, noting dduln⁡(sin⁡u+cos⁡u)=cos⁡u−sin⁡usin⁡u+cos⁡u\dfrac{d}{du}\ln(\sin u+\cos u)=\dfrac{\cos u-\sin u}{\sin u+\cos u}:
I=2[u2−12ln⁡(sin⁡u+cos⁡u)]0π/4=2[π8−12ln⁡2+12ln⁡1]=π4−ln⁡2.I=2\left[\dfrac{u}{2}-\dfrac12\ln(\sin u+\cos u)\right]_{0}^{\pi/4}=2\left[\dfrac{\pi}{8}-\dfrac12\ln\sqrt{2}+\dfrac12\ln 1\right]=\dfrac{\pi}{4}-\ln\sqrt{2}.
 Correct answer: (3)
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