Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals3 Marks
Question
If $\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8},$ find the value of a.
✓
Answer
$\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8}$ $\Rightarrow\frac{1}{2}\tan^{-1}\Big[\frac{\text{x}}{2}\Big]^{\text{a}}_0=\frac{\pi}{8}$$\bigg[\int\frac{1}{\text{a}^2+\text{x}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\bigg]$ $\Rightarrow\frac{1}{2}\Big(\tan^{-1}\frac{\text{a}}{2}-\tan^{-1}0\Big)=\frac{\pi}{8}$ $\Rightarrow\tan^{-1}\frac{\text{a}}{2}-0=\frac{\pi}{4}$ $\Rightarrow\tan^{-1}\frac{\text{a}}{2}=\frac{\pi}{4}$ $\Rightarrow\frac{\text{a}}{2}=\tan\frac{\pi}{4}=1$ $\Rightarrow\text{a}=2$ Thus, the value of a is 2
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