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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate $ \int \limits _0^1 \tan ^{-1}(x^2+1-x)dx$

$(a)\;\frac{\pi}{2}+1 \\(b)\;\frac{\pi}{2}\\(c)\;\frac{\pi}{2}-1 \\ (d)\;\pi $

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1 Answer

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$\int \limits_0^1 \large\frac{\pi}{2}$$ - \cot ^{-1} (1+x^2-x)$
=> $\int\limits_0^1 \large\frac{\pi}{2}$$ - \cot ^{-1} (1+x(-1+x)] dx$
=> $\int\limits_0^1 \large\frac{\pi}{2}$$ - \tan ^{-1} \bigg(\large\frac{1}{1+x(1-x)}\bigg)$$dx$
=> $\int\limits_0^1 \large\frac{\pi}{2}$$ - \tan ^{-1} \bigg(\large\frac{+x-(-1+x)}{1+x(-1+x)}\bigg)$$dx$
=> $\int \limits _0^1 \large\frac{\pi}{2}$$ - \tan ^{-1} [(x-1) - x/1+x(x-1)]dx$
=> $\int \limits _0^1 \large\frac{\pi}{2}$$ + \tan ^{-1} (x-1) - \tan ^{-1}(x)dx$
=> $\int \limits _0^1 \large\frac{\pi}{2}$$+ \int \limits_0^1 \tan ^{-1} (x-1) dx -\int \limits_0^1 \tan (x) dx $
=> $\int \limits _0^1 \large\frac{\pi}{2}$$dx+ \int \limits_0^1 \tan ^{-1} (x-1) dx -\int \limits_0^1 \tan^{-1} x dx $
$f(x)=f(a-x)$ by this property
=> $\int \limits _0^1 \large\frac{\pi}{2}$$+ \int \limits_0^1 \tan ^{-1} (1-x-1) dx -\int \tan^{-1} (x) dx $
=> $\int \limits _0^1 \large\frac{\pi}{2}$$+ 2 \tan ^{-1}(x) dx$
By applying the limits we get,
$\large\frac{\pi}{2}$$+1$
Hence a is the correct answer.
answered Dec 30, 2013 by meena.p
 
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