Definite Integration Question 70
Question 70
- 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, $I_{1}=\int_{0}^{1} \tan ^{-1} x d x=\left[x \tan ^{-1} x\right]{0}^{1}-\int{0}^{1} \frac{x d x}{1+x^{2}}$
$$ \begin{gathered} =\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right]{0}^{1}=\frac{\pi}{4}-\frac{1}{2} \log 2 \ \therefore \int{0}^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x=\frac{\pi}{2}-2 \frac{\pi}{4}-\frac{1}{2} \log 2=\log 2 \end{gathered} $$
