The area of the region between the curves y = \sqrt{\frac{1+sinx}{cosx}} and y = \sqrt{\frac{1-sinx}{cosx} bounded by the lines x = 0 and x = \pi/4 is [JEE Adv 2008 P2]

Correct Answer is : √2−1∫04t(1+t2)√1−t2

\int^{\sqrt2-1}_{0} t/{(1+t^2)\sqrt{1-t^2}

\int^{\sqrt2-1}_{0} {4t}/{(1+t^2)\sqrt{1-t^2}

\int^{\sqrt2+1}_{0} {4t}/{(1+t^2)\sqrt{1-t^2}

\int^{\sqrt2+1}_{0} t/{(1+t^2)\sqrt{1-t^2}

Application of Integration

Section: Mathematics

Question : 27 of 28

Marks: +1, -0

The area of the region between the curves
y = √

1+sinx

cosx

and y = √

1−sinx

cosx

bounded by the lines x = 0 and x =

[JEE Adv 2008 P2]

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