The area of the region between the curves y= (1+ x)/( x) and y= (1- x)/( x) bounded by the lines x=0 and x=(π)/(4) is

Correct answer: B. ∫_0^ 2 -1 4 t (1+t^2) 1-t^2 d t aligned & Required area =∫_0^ π / 4 ( (1+ x)/( x) - (1- x)/( x) ) d x [ (1+ x)/( x)>(1- x)/( x)>0] & .=∫_0^ π / 4 (1+(2 (x)/(2))/(1+ ^2 (x)/(2)))/((1- ^2 (x)/(2))/(1+ ^2 (x)/(2))) (1-(2 (x)/(2))/(1+ ^2 (x)/(2)))/((1- ^2 (x)/(2))/(1+ ^2 (x)/(2))) ) d x & .=∫_0^ π / 4 (1+ (x)/(2))/(1- (x)/(2)) - (1- (x)/(2))/(1+ (x)/(2)) ) d x & aligned =∫_0^ π / 4 1+ (x)/(2)-1+ (x)/(2) 1- ^2 (x)/(2) d x=∫_0^ π / 4 2 (x)/(2) 1- ^2 (x)/(2) d x Put (x)/(2)=t aligned (1)/(2) ^2 (x)/(2) d x & =d t Required area & =∫_0^ (π)/(8) 4 t d t (1+t^2) 1-t^2 & =∫_0^ 2 -1 4 t (1+t^2) 1-t^2 d t [ astan (π)/(8)= 2 -1] aligned

JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Mathematics
Chapter: Area Under Curves

JEE Advanced 2008 · Paper 2 · Mathematics · Area Under Curves

The area of the region between the curves y= (1+ x)/( x) and y= (1-...

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}}

bounded by the lines

x=0

and

x=\frac{\pi}{4}

Options

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 $\int_0^{\sqrt{2}+1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

Correct answer

B. $\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

Step-by-step solution

\begin{aligned} & \text { Required area }=\int_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x \quad\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right] \\ & \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}} \sqrt{\frac{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}}\right) d x \\ & \left.=\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}}}\right) d x \\ & \end{aligned}

=\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

Put

\tan \frac{x}{2}=t

\begin{aligned} \Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x & =d t \\ \therefore \text { Required area } & =\int_0^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^2\right) \sqrt{1-t^2}} \\ & =\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \quad\left[\text { astan } \frac{\pi}{8}=\sqrt{2}-1\right] \end{aligned}

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