AP EAMCET · Maths · Definite Integration
\(\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x=\)
- A \(\frac{\pi-2}{2}\)
- B \(\frac{\pi+2}{2}\)
- C \(\frac{\pi(\pi+2)}{2}\)
- D \(\frac{\pi(\pi-2)}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi(\pi-2)}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & I=\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x \\ & \qquad \because \text { using } \\ & \qquad \int_0^a f(x) d x=\int_0^a f(a-x) d x\end{aligned}\) \(=\int_0^\pi \frac{-(\pi-x) \tan (x)}{-\sec x-\tan x} d x\) Adding Eqs. (i) and (ii)…
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