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