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 $x=\frac{\pi}{4}$ is (a) $\int_{0}^{\sqrt{2}-1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$ (b) $\int_{0}^{\sqrt{2}-1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$ (c) $\int_{0}^{\sqrt{2}+1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$ (d) $\int_{0}^{\sqrt{2}+1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$
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Solution:
- Required area $=\int_{0}^{\pi / 4} \sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}} d x$
$$ \begin{aligned} & \because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0 \end{aligned} $$
$$ \begin{aligned} & =\int_{0}^{\pi / 4} \sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}} d x \ & =\int_{0}^{\pi / 4} \frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-\tan ^{2} \frac{x}{2}}} d x=\int_{0}^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^{2} \frac{x}{2}}} d x \end{aligned} $$
Put $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t=\int_{0}^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}}$
$$ \text { As } \quad \int_{0}^{\sqrt{2}-1} \frac{4 t d t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} \quad\left[\because \tan \frac{\pi}{8}=\sqrt{2}-1\right] $$
