SATHEE: Definite Integration Question 60

Definite Integration Question 60

Question 60

The value of $\int_{π / 4}^{3 π / 4} \frac{x}{1 + \sin x} d x \dots \dots$ .

$(1993, 2 M)$

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Solution:

Let $I = \int_{π / 4}^{3 π / 4} \frac{x}{1 + \sin x} d x$

$\begin{aligned} \Rightarrow I = \int_{π / 4}^{3 π / 4} \frac{\frac{π}{4} + \frac{3 π}{4} - x}{1 + \sin \frac{π}{4} + \frac{3 π}{4} - x} d x & [∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] \end{aligned}$

$= \int_{π / 4}^{3 π / 4} \frac{π - x}{1 + \sin (π - x)} d x$

$= \int_{π / 4}^{3 π / 4} \frac{π}{1 + \sin x} d x - \int_{π / 4}^{3 π / 4} \frac{x}{1 + \sin x} d x$

$= π \int_{π / 4}^{3 π / 4} \frac{d x}{1 + \sin x} - I$

[from Eq. (i)]

$= \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{d x}{(1 + \sin x)}$

$= \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{(1 - \sin x)}{(1 + \sin x) (1 - \sin x)} d x$

$= \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{(1 - \sin x)}{1 - \sin^{2} x} d x$

$= \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x} d x$

$= \frac{π}{2} \int_{π / 4}^{3 π / 4} (\sec^{2} x - \sec x \cdot \tan x) d x$

$= \frac{π}{2} [\tan x - \sec x]_{π / 4}^{3 π / 4}$

$= \frac{π}{2} [- 1 - 1 - (- \sqrt{2} - \sqrt{2})]$

$= \frac{π}{2} (- 2 + 2 \sqrt{2}) = π (\sqrt{2} - 1)$

Definite Integration Question 60

Question 60

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Definite Integration Question 60

Question 60

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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