Definite Integration Question 72
Question 72
- Evaluate the definite integral
$$ \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1-x^{4}} \cos ^{-1} \frac{2 x}{1+x^{2}} d x $$
$(1995,5 M)$
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Solution:
- Let $I=\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1-x^{4}} \cos ^{-1} \frac{2 x}{1+x^{2}} d x$
Put $x=-y \Rightarrow d x=-d y$
$\therefore \quad I=\int_{1 / \sqrt{3}}^{-1 / \sqrt{3}} \frac{y^{4}}{1-y^{4}} \cos ^{-1} \frac{-2 y}{1+y^{2}}(-1) d y$ Now, $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$ for $-1 \leq x \leq 1$.
$$ \begin{aligned} \therefore \quad I & =\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1-y^{4}} \pi-\cos ^{-1} \frac{2 y}{1+y^{2}} d y \ & =\pi \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1-y^{4}} d y-\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1-y^{4}} \cos ^{-1} \frac{2 y}{1+y^{2}} d y \ & =\pi \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1-x^{4}} d x-\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1-x^{4}} \cos ^{-1} \frac{2 x}{1+x^{2}} d x \ \Rightarrow \quad I & =\pi \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1-x^{4}} d x-I \ \Rightarrow \quad 2 I= & \quad \text { [from Eq. (i)] } \ = & 1 / \sqrt{3} \frac{x^{4}}{1-x^{4}} d x=\pi \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}}-1+\frac{1}{1-x^{4}} d x \ = & \quad 2 I=-\pi[x]{-1 / \sqrt{3}}^{1 / \sqrt{3}}+\pi I{1}, \text { where } I_{1}=\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{d x}{1-x^{4}} \ \Rightarrow \quad & \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\pi I_{1}=-\frac{2 \pi}{\sqrt{3}}+\pi I_{1} \end{aligned} $$
Now, $I_{1}=\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{d x}{1-x^{4}}=2 \int_{0}^{1 / \sqrt{3}} \frac{d x}{1-x^{4}}$
[since, the integral is an even function]
$=\int_{0}^{1 / \sqrt{3}} \frac{1+1+x^{2}-x^{2}}{\left(1-x^{2}\right)\left(1+x^{2}\right)} d x$
$=\int_{0}^{1 / \sqrt{3}} \frac{1}{1-x^{2}} d x+\int_{0}^{1 / \sqrt{3}} \frac{1}{1+x^{2}} d x$
$=\int_{0}^{1 / \sqrt{3}} \frac{1}{(1-x)(1+x)} d x+\int_{0}^{1 / \sqrt{3}} \frac{1}{\left(1+x^{2}\right)} d x$
$=\frac{1}{2} \int_{0}^{1 / \sqrt{3}} \frac{1}{1-x} d x+\frac{1}{2} \int_{0}^{1 / \sqrt{3}} \frac{1}{1+x} d x+\int_{0}^{1 / \sqrt{3}} \frac{1}{1+x^{2}} d x$
$=-\frac{1}{2} \ln |1-x|+\frac{1}{2} \ln |1+x|+\tan ^{-1} x_{0}^{1 / \sqrt{3}}$
$=\frac{1}{2} \ln \frac{1+x}{1-x}{ }{0}^{1 / \sqrt{3}}+\left[\tan ^{-1} x\right]{0}^{1 / \sqrt{3}}$
$=\frac{1}{2} \ln \frac{1+1 / \sqrt{3}}{1-1 / \sqrt{3}}+\tan ^{-1} \frac{1}{\sqrt{3}}$
$=\frac{1}{2} \ln \frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\pi}{6}=\frac{1}{2} \ln \frac{(\sqrt{3}+1)^{2}}{3-1}+\frac{\pi}{6}$
$=\frac{1}{2} \ln (2+\sqrt{3})+\frac{\pi}{6}$
$\therefore 2 I=\frac{-2 \pi}{\sqrt{3}}+\frac{\pi}{2} \ln (2+\sqrt{3})+\frac{\pi^{2}}{6}$
$=\frac{\pi}{6}[\pi+3 \ln (2+\sqrt{3})-4 \sqrt{3}]$
$\Rightarrow I=\frac{\pi}{12}[\pi+3 \ln (2+\sqrt{3})-4 \sqrt{3}]$
Alternate Solution
Since, $\quad \cos ^{-1} y=\frac{\pi}{2}-\sin ^{-1} y$
$\therefore \cos ^{-1} \frac{2 x}{1+x^{2}}=\frac{\pi}{2}-\sin ^{-1} \frac{2 x}{1+x^{2}}=\frac{\pi}{2}-2 \tan ^{-1} x$
$I=\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{\pi}{2} \cdot \frac{x^{4}}{1-x^{4}}-\frac{x^{4}}{1-x^{4}} 2 \tan ^{-1} x d x$
$\because \frac{x^{4}}{1-x^{4}} 2 \tan ^{-1} x$ is an odd function
$\therefore \quad I=2 \cdot \frac{\pi}{2} \int_{0}^{\frac{1}{\sqrt{3}}}-1+\frac{1}{1-x^{4}} \quad d x+0$
$=\frac{\pi}{2} \int_{0}^{1 / \sqrt{3}}-2+\frac{1}{1-x^{2}}+\frac{1}{1+x^{2}} d x$
$=\frac{\pi}{2}-2 x+\frac{1}{2 \cdot 1} \log \frac{1+x}{1-x}+\tan ^{-1} x_{0}^{1 / \sqrt{3}}$
$=\frac{\pi}{2}-\frac{2}{\sqrt{3}}+\frac{1}{2} \log \frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\pi}{6}$
$=\frac{\pi}{12}[\pi+3 \log (2+\sqrt{3})-4 \sqrt{3}]$
