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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercise : The value of $\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}\ $ is
  • A
    $0$
  • B
    $2$
  • ✓
    $\pi$
  • D
    $1$
✓

Answer

Correct option: C.
$\pi$
Let $\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}^{3}\text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}\cos\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\tan^{5}\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}1\text{dx}$
$\Rightarrow\ \ \text{I}=0+0+0+\text{(x)}^{\frac{\pi}{2}}_{\frac{-\pi}{2}}=\bigg(\frac{-\pi}{2}\bigg)=\pi$

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