Indefinite Integration 1 Question 70
71. Prove that $\int _0^{1} \tan ^{-1} \frac{1}{1-x+x^{2}} d x=2 \int _0^{1} \tan ^{-1} x d x$.
Hence or otherwise, evaluate the integral
$$ \int _0^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x \text {. } $$
$(1998,8$ M)
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Solution:
- $\int _0^{1} \tan ^{-1} \frac{1}{1-x+x^{2}} d x=\int _0^{1} \tan ^{-1} \frac{1-x+x}{1-x(1-x)} d x$
$$ \begin{aligned} & =\int _0^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1} x\right] d x \\ & =\int _0^{1} \tan ^{-1}[1-(1-x)] d x+\int _0^{1} \tan ^{-1} x d x \\ & =2 \int _0^{1} \tan ^{-1} x d x \quad \because \int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x \end{aligned} $$
Now, $\int _0^{1} \tan ^{-1} \frac{1}{1-x+x^{2}} d x$
$$ \begin{aligned} &=\int _0^{1} \frac{\pi}{2}-\cot ^{-1} \frac{1}{1-x+x^{2}} d x \\ &= \frac{\pi}{2}-\int _0^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x \\ & \therefore \quad \int _0^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x=\frac{\pi}{2}-\int _0^{1} \tan ^{-1} \frac{1}{\left(1-x+x^{2}\right)} d x \\ &=\frac{\pi}{2}-2 I _1 \end{aligned} $$
where, $I _1=\int _0^{1} \tan ^{-1} x d x=\left[x \tan ^{-1} x\right] _0^{1}-\int _0^{1} \frac{x d x}{1+x^{2}}$
$$ \begin{gathered} =\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right] _0^{1}=\frac{\pi}{4}-\frac{1}{2} \log 2 \\ \therefore \int _0^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x=\frac{\pi}{2}-2 \frac{\pi}{4}-\frac{1}{2} \log 2=\log 2 \end{gathered} $$
