MHT CET · Maths · Definite Integration
If \(\int_{o}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}\), then \(a=\)
- A \(\frac{1}{2}\)
- B 2
- C \(\frac{1}{4}\)
- D 1
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
We have \(\int_{0}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}\)
\(\therefore \frac{\pi}{8}=\frac{1}{4} \int_{0}^{a} \frac{d x^{2}+\left(\frac{1}{2}\right)^{2}}{2}=\frac{1}{4} \frac{1}{\left(\frac{1}{2}\right)}\left[\tan ^{-1} \frac{x}{\left(\frac{1}{2}\right)}\right]_{0}^{a}\)
\(\quad=\frac{1}{2}\left[\tan ^{-1} 2 a-\tan ^{-1} 0\right]\)
\(\therefore \frac{\pi}{8}=\frac{1}{2} \tan ^{-1} 2 a \Rightarrow \frac{\pi}{4}=\tan ^{-1} 2 a \Rightarrow 2 a=\tan \frac{\pi}{4}=\) \(1 \Rightarrow a=\frac{1}{2}\)
\(\therefore \frac{\pi}{8}=\frac{1}{4} \int_{0}^{a} \frac{d x^{2}+\left(\frac{1}{2}\right)^{2}}{2}=\frac{1}{4} \frac{1}{\left(\frac{1}{2}\right)}\left[\tan ^{-1} \frac{x}{\left(\frac{1}{2}\right)}\right]_{0}^{a}\)
\(\quad=\frac{1}{2}\left[\tan ^{-1} 2 a-\tan ^{-1} 0\right]\)
\(\therefore \frac{\pi}{8}=\frac{1}{2} \tan ^{-1} 2 a \Rightarrow \frac{\pi}{4}=\tan ^{-1} 2 a \Rightarrow 2 a=\tan \frac{\pi}{4}=\) \(1 \Rightarrow a=\frac{1}{2}\)
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