AP EAMCET · Maths · Definite Integration
\(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x\) is equal to
- A \(0\)
- B \(\frac{\pi}{4}\)
- C \(\frac{-\pi}{4}\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{-\pi / 4}^{\pi / 4} x \tan \left(1+x^2\right) d x\) Here, \(f(x)=x \tan \left(1+x^2\right)\) \(f(-x)=-x \tan \left(1+x^2\right)=-f(x)\) \(f(x)\) is an odd function. \(\therefore \quad I=0\)
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