SATHEE: Indefinite Integration 1 Question 73

Indefinite Integration 1 Question 73

74. 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 $x = - y \Rightarrow d x = - d y$

$∴ I = \int_{1 / \sqrt{3}}^{- 1 / \sqrt{3}} \frac{y^{4}}{1 - y^{4}} \cos^{- 1} \frac{- 2 y}{1 + y^{2}} (- 1) d y$ Now, $\cos^{- 1} (- x) = π - \cos^{- 1} x$ for $- 1 \leq x \leq 1$ .

$\begin{aligned} ∴ I & = \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1 - y^{4}} π - \cos^{- 1} \frac{2 y}{1 + y^{2}} d y \\ = π \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1 - y^{4}} d y - \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{y^{4}}{1 - y^{4}} \cos^{- 1} \frac{2 y}{1 + y^{2}} d y \\ = π \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1 - x^{4}} d x - \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1 - x^{4}} \cos^{- 1} \frac{2 x}{1 + x^{2}} d x \\ \Rightarrow I & = π \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^{4}}{1 - x^{4}} d x - I \\ \Rightarrow 2 I = & [from Eq. (i)] \\ = & 1 / \sqrt{3} \frac{x^{4}}{1 - x^{4}} d x = π \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} - 1 + \frac{1}{1 - x^{4}} d x \\ = & 2 I = - π [x]_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} + π I_{1}, where I_{1} = \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{d x}{1 - x^{4}} \\ \Rightarrow & \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} + π I_{1} = - \frac{2 π}{\sqrt{3}} + π I_{1} \end{aligned}$

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]

$= \int_{0}^{1 / \sqrt{3}} \frac{1 + 1 + x^{2} - x^{2}}{(1 - x^{2}) (1 + x^{2})} d x$

$= \int_{0}^{1 / \sqrt{3}} \frac{1}{1 - x^{2}} d x + \int_{0}^{1 / \sqrt{3}} \frac{1}{1 + x^{2}} d x$

$= \int_{0}^{1 / \sqrt{3}} \frac{1}{(1 - x) (1 + x)} d x + \int_{0}^{1 / \sqrt{3}} \frac{1}{(1 + x^{2})} d x$

$= \frac{1}{2} \int_{0}^{1 / \sqrt{3}} \frac{1}{1 - x} d x + \frac{1}{2} \int_{0}^{1 / \sqrt{3}} \frac{1}{1 + x} d x + \int_{0}^{1 / \sqrt{3}} \frac{1}{1 + x^{2}} d x$

$= - \frac{1}{2} \ln | 1 - x | + \frac{1}{2} \ln | 1 + x | + \tan^{- 1} x_{0}^{1 / \sqrt{3}}$

$= \frac{1}{2} \ln \frac{1 + x}{1 - x}_{0}^{1 / \sqrt{3}} + {[\tan^{- 1} x]}_{0}^{1 / \sqrt{3}}$

$= \frac{1}{2} \ln \frac{1 + 1 / \sqrt{3}}{1 - 1 / \sqrt{3}} + \tan^{- 1} \frac{1}{\sqrt{3}}$

$= \frac{1}{2} \ln \frac{\sqrt{3} + 1}{\sqrt{3} - 1} + \frac{π}{6} = \frac{1}{2} \ln \frac{(\sqrt{3} + 1)^{2}}{3 - 1} + \frac{π}{6}$

$= \frac{1}{2} \ln (2 + \sqrt{3}) + \frac{π}{6}$

$∴ 2 I = \frac{- 2 π}{\sqrt{3}} + \frac{π}{2} \ln (2 + \sqrt{3}) + \frac{π^{2}}{6}$

$= \frac{π}{6} [π + 3 \ln (2 + \sqrt{3}) - 4 \sqrt{3}]$

$\Rightarrow I = \frac{π}{12} [π + 3 \ln (2 + \sqrt{3}) - 4 \sqrt{3}]$

Alternate Solution

Since, $\cos^{- 1} y = \frac{π}{2} - \sin^{- 1} y$

$∴ \cos^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{2} - \sin^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{2} - 2 \tan^{- 1} x$

$I = \int_{- 1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{π}{2} \cdot \frac{x^{4}}{1 - x^{4}} - \frac{x^{4}}{1 - x^{4}} 2 \tan^{- 1} x d x$

$∵ \frac{x^{4}}{1 - x^{4}} 2 \tan^{- 1} x$ is an odd function

$∴ I = 2 \cdot \frac{π}{2} \int_{0}^{\frac{1}{\sqrt{3}}} - 1 + \frac{1}{1 - x^{4}} d x + 0$

$= \frac{π}{2} \int_{0}^{1 / \sqrt{3}} - 2 + \frac{1}{1 - x^{2}} + \frac{1}{1 + x^{2}} d x$

$= \frac{π}{2} - 2 x + \frac{1}{2 \cdot 1} \log \frac{1 + x}{1 - x} + \tan^{- 1} x_{0}^{1 / \sqrt{3}}$

$= \frac{π}{2} - \frac{2}{\sqrt{3}} + \frac{1}{2} \log \frac{\sqrt{3} + 1}{\sqrt{3} - 1} + \frac{π}{6}$

$= \frac{π}{12} [π + 3 \log (2 + \sqrt{3}) - 4 \sqrt{3}]$

Indefinite Integration 1 Question 73

74. Evaluate the definite integral

Solution:

Alternate Solution

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Indefinite Integration 1 Question 73

74. Evaluate the definite integral

Solution:

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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