106nope [2] ans : -pi/2
Q1. \int_{-1}^{1}{\frac{d(tan^{-1}\frac{1}{x})}{dx}dx}
Q2. \int_{-1}^{3}{(tan^{-1}\frac{x^2+1}{x}+tan^{-1}\frac{x}{x^2+1})dx}
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16 Answers
1for second one
\int \tan^{-1} \frac{x^2 + 1}{x}+ cot^{-1}\frac{x^2 + 1}{x} dx
\int \frac{\pi }{2} dx \rightarrow \frac{\pi }{2}\int_{-1}^{3}{dx}= \frac{\pi }{2}\left(3- ( -1 ) \right) \rightarrow \frac{\pi }{2} .4\rightarrow 2\pi
24ya soln 2 is rite
@asish...
have u solved 1 ???i cant understand how ans is -pi/2
@SR
i have time and again critisiced the use of this stuff.....plz dont post data from mathematica ...I hope u take it in right spirit[1][6]
106for 2. every one making same mistake (i think 99.9% sure)
for 1. i think the book (RD Sharma) is geniune in giving answer as -pi/2
106so, i presume answer given in the book is wrong??? (given: 2pi)
1\tan^{-1} \frac{1}{x} = cot^{-1}x( in ,this ,case, it ,is, true )
cot ^{-1}x ( applying limits) = cot^{-1}( 1 ) - \left(cot ^{-1} ( -1 ) \right) =\frac{\pi }{2} - \left(\pi - cot^{-1} 1\right)\rightarrow \frac{\pi }{4} -\pi +\frac{\pi }{4} = \frac{\pi }{2}-\pi = -\frac{\pi }{2}
1Q1) \int_{-1}^{0}{}(cot^{-1}x-\pi )dx+\int_{0}^{1}{cot^{-1}x}dx
isse -\pi aa raha hai...
in career point material given that
cot-1x = tan-1(1x), x \epsilon (0 , \infty)
= \pi + tan-1(1x) , x \epsilon ( -- \infty , 0)