Indefinite Integration 1 Question 4
5. The value of the integral $\int _0^{1} x \cot ^{-1}\left(1-x^{2}+x^{4}\right) d x$ is
(a) $\frac{\pi}{4}-\frac{1}{2} \log _e 2$
(b) $\frac{\pi}{2}-\frac{1}{2} \log _e 2$
(c) $\frac{\pi}{4}-\log _e 2$
(d) $\frac{\pi}{2}-\log _e 2$
(2019 Main, 9 April II)
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Solution:
- Let $I=\int _0^{1} x \cot ^{-1}\left(1-x^{2}+x^{4}\right) d x$
Now, put $x^{2}=t \Rightarrow 2 x d x=d t$
Lower limit at $x=0, t=0$
Upper limit at $x=1, t=1$
$\therefore I=\frac{1}{2} \int _0^{1} \cot ^{-1}\left(1-t+t^{2}\right) d t$
$$ \begin{aligned} & \quad=\frac{1}{2} \int _0^{1} \tan ^{-1} \frac{1}{1-t+t^{2}} d t \quad \because \cot ^{-1} x=\tan ^{-1} \frac{1}{x} \\ & =\frac{1}{2} \int _0^{1} \tan ^{-1} \frac{t-(t-1)}{1+t(t-1)} d t \\ & =\frac{1}{2} \int _0^{1}\left(\tan ^{-1} t-\tan ^{-1}(t-1) d t\right. \\ & \because \tan ^{-1} \frac{x-y}{1+x y}=\tan ^{-1} x-\tan ^{-1} y \\ & \because \int _0^{1} \tan ^{-1}(t-1) d t=\int _0^{1} \tan ^{-1}(1-t-1) d t=-\int _0^{1} \tan ^{-1}(t) d t \end{aligned} $$
because $\int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x$
So, $I=\frac{1}{2} \int _0^{1}\left(\tan ^{-1} t+\tan ^{-1} t\right) d t$
$$ =\int _0^{1} \tan ^{-1} t d t=\left[t \tan ^{-1} t\right] _0^{1}-\int _0^{1} \frac{t}{1+t^{2}} d t $$
[by integration by parts method]
$$ =\frac{\pi}{4}-\frac{1}{2}\left[\log _e\left(1+t^{2}\right)\right] _0^{1}=\frac{\pi}{4}-\frac{1}{2} \log _e 2 $$
