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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} d x=$
  • A
    $\frac{\pi}{4}+\frac{1}{2} \log 2$
  • B
    $\frac{\pi}{4}+\log 2$
  • ✓
    $\frac{\pi}{4}-\frac{1}{2} \log 2$
  • D
    $\frac{\pi}{4}-\log 2$
✓

Answer

Correct option: C.
$\frac{\pi}{4}-\frac{1}{2} \log 2$
(C)
$\int_0^{\frac{\pi}{2}} \frac{\cos x}{1+\cos x+\sin x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{2 \cos ^2(x / 2)+2 \sin (x / 2) \cos (x / 2)} d x$
$\begin{array}{l}=\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1-\tan ^2(x / 2)}{1+\tan (x / 2)} d x \\ =\frac{1}{2} \int_0^{\frac{\pi}{2}}\left[1-\tan \left(\frac{x}{2}\right)\right] d x \\ =\frac{1}{2}\left[x+2 \log \left|\cos \left(\frac{x}{2}\right)\right|\right]_0^{\frac{\pi}{2}} \\ =\frac{\pi}{4}+\log \frac{1}{\sqrt{2}} \\ =\frac{\pi}{4}-\frac{1}{2} \log 2\end{array}$

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