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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is :
  • A
    $\frac{\pi^2}{4}$
  • B
    $\frac{\pi^2}{2}$
  • ✓
    $\frac{3\pi^2}{2}$
  • D
    $\frac{\pi^2}{2}$
✓

Answer

Correct option: C.
$\frac{3\pi^2}{2}$
$\int\limits^{2\pi}_0\sqrt{1+\sin\frac{\text{x}}{2}}\text{dx}$
$=\int\limits^{2\pi}_0\sqrt{\sin^2\frac{\text{x}}{4}+\cos^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{dx}$
$=\int\limits^{2\pi}_0\Big(\sin\frac{\text{x}}{4}+\cos\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{-\cos^\frac{\text{x}}{4}}{\frac{1}{4}}+\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big[\frac{\text{x}}{4}-\cos\frac{\text{x}}{4}\Big]^{2\pi}_0$
$=4\Big[\sin\frac{2\pi}{4}-\cos\frac{2\pi}{4}-\sin0+\cos0\Big]$
$=4\Big[\sin\frac{\pi}{2}-\cos\frac{\pi}{2}-0+1\Big]$
$=4\big[1-0-0+1\big]$
$=4\times2$
$=8$

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