SATHEE: Definite Integration Question 71

Definite Integration Question 71

Question 71

Integrate $\int_{0}^{π / 4} \log (1 + \tan x) d x$ .

(1997C, 2M)

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Solution:

Let $I = \int_{0}^{π / 4} \log (1 + \tan x) d x$

$\begin{aligned} I = \int_{0}^{π / 4} \log (1 + \tan (\frac{π}{4} - x)) d x ∴ & I = \int_{0}^{π / 4} \log 1 + \frac{1 - \tan x}{1 + \tan x} d x \end{aligned}$

$\begin{aligned} \int_{0}^{π / 4} \log \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} d x & I = \int_{0}^{π / 4} \log \frac{2}{1 + \tan x} d x \Rightarrow I = \int_{0}^{π / 4} \log 2 d x - I & \Rightarrow 2 I = \frac{π}{4} \log 2 \Rightarrow I = \frac{π}{8} (\log 2) & 73. Let I = \int_{- π}^{π} \frac{2 x (1 + \sin x)}{1 + \cos^{2} x} d x & I = \int_{- π}^{π} \frac{2 x}{1 + \cos^{2} x} d x + \int_{- π}^{π} \frac{2 x \sin x}{1 + \cos^{2} x} d x & I = I_{1} + I_{2} \end{aligned}$

Now, $I_{1} = \int_{- π}^{π} \frac{2 x}{1 + \cos^{2} x} d x$

Let $f (x) = \frac{2 x}{1 + \cos^{2} x}$

$\Rightarrow f (- x) = \frac{- 2 x}{1 + \cos^{2} (- x)} = \frac{- 2 x}{1 + \cos^{2} x} = - f (x)$

$\Rightarrow f (- x) = - f (x)$ which shows that $f (x)$ is an odd function.

$\begin{aligned} ∴ & I_{1} & = 0 Again, let & g (x) & = \frac{2 x \sin x}{1 + \cos^{2} x} \end{aligned}$

$\Rightarrow g (- x) = \frac{2 (- x) \sin (- x)}{1 + \cos^{2} (- x)} = \frac{2 x \sin x}{1 + \cos^{2} x} = g (x)$

$\Rightarrow g (- x) = g (x)$ which shows that $g (x)$ is an even function.

$∴ I_{2} = \int_{- π}^{π} \frac{2 x \sin x}{1 + \cos^{2} x} d x = 2 \cdot 2 \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$

$\begin{aligned} = 4 \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + [\cos (π - x)]^{2}} d x = 4 \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} d x & = 4 \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} d x - 4 \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x \end{aligned}$

$\Rightarrow I_{2} = 4 π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} d x - I_{2}$

$\Rightarrow 2 I_{2} = 4 π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} d x$

Put $\cos x = t \Rightarrow - \sin x d x = d t$

$∴ I_{2} = - 2 π \int_{1}^{- 1} \frac{d t}{1 + t^{2}} = 2 π \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = 4 π \int_{0}^{1} \frac{d t}{1 + t^{2}}$

$= 4 π {[\tan^{- 1} t]}_{0}^{1} = 4 π [\tan^{- 1} 1 - \tan^{- 1} 0]$

$= 4 π (π / 4 - 0) = π^{2}$

$∴ I = I_{1} + I_{2} = 0 + π^{2} = π^{2}$

Definite Integration Question 71

Question 71

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Definite Integration Question 71

Question 71

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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