Question
Download Solution PDF\(\rm \int^1_0 \frac {\tan^{-1} x}{1 + x^2}dx\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
\( \rm \int x^{n}dx = \frac{x^{n+1}}{n+1} + c\)
गणना:
माना कि I = \(\rm \int^1_0 \frac {\tan^{-1} x}{1 + x^2}dx\) है।
माना कि tan-1 x = t है।
दोनों पक्षों का अवकलन करने पर, हमें निम्न प्राप्त होता है
⇒ \(\rm \frac{1}{1+x^2} dx = dt\)
|
x |
0 |
1 |
|
t |
0 |
π/4 |
अब, I \(= \rm \int _0^{\pi/4} t\;dt\)
\( = \rm [\frac{t^2}{2}]_0^{\pi/4}\)
\(= \rm \frac{1}{2}[\frac{\pi^2}{16}-0]\)
= \(\rm \frac {\pi^2}{32}\)
Last updated on Jul 1, 2025
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