SATHEE: Area Question 23

Area Question 23

Question 23

The area of the region between the curves $y = \sqrt{\frac{1 + \sin x}{\cos x}}$ and $y = \sqrt{\frac{1 - \sin x}{\cos x}}$ and bounded by the lines $x = 0$ and $x = \frac{π}{4}$ is (a) $\int_{0}^{\sqrt{2} - 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$ (b) $\int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$ (c) $\int_{0}^{\sqrt{2} + 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$ (d) $\int_{0}^{\sqrt{2} + 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

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Solution:

Required area $= \int_{0}^{π / 4} \sqrt{\frac{1 + \sin x}{\cos x}} - \sqrt{\frac{1 - \sin x}{\cos x}} d x$

$\begin{aligned} ∵ \frac{1 + \sin x}{\cos x} > \frac{1 - \sin x}{\cos x} > 0 \end{aligned}$

$\begin{aligned} = \int_{0}^{π / 4} \sqrt{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}} - \sqrt{\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}} d x & = \int_{0}^{π / 4} \frac{1 + \tan \frac{x}{2} - 1 + \tan \frac{x}{2}}{\sqrt{1 - \tan^{2} \frac{x}{2}}} d x = \int_{0}^{π / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1 - \tan^{2} \frac{x}{2}}} d x \end{aligned}$

Put $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t = \int_{0}^{\tan \frac{π}{8}} \frac{4 t d t}{(1 + t^{2}) \sqrt{1 - t^{2}}}$

$As \int_{0}^{\sqrt{2} - 1} \frac{4 t d t}{(1 + t^{2}) \sqrt{1 - t^{2}}} [∵ \tan \frac{π}{8} = \sqrt{2} - 1]$

Area Question 23

Question 23

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Area Question 23

Question 23

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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