Integral Calculus
Area Between Curves
MMTS_Full_Test_16
Grade 12
Question:
The area of the region between $y=\sqrt{\dfrac{1+\sin x}{\cos x}}$ and $y=\sqrt{\dfrac{1-\sin x}{\cos x}}$ bounded by $x=0$ and $x=\pi/4$ is
$\int_0^{\sqrt{2}-1}\dfrac{t}{(1+t^2)\sqrt{1-t^2}}dt$
$\int_0^{\sqrt{2}-1}\dfrac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
$\int_0^{\sqrt{2}+1}\dfrac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
$\int_0^{\sqrt{2}+1}\dfrac{t}{(1+t^2)\sqrt{1-t^2}}dt$
Step-by-Step Solution
Key Concept: Simplify the functions using half-angle; substitute $t=\tan(x/2)$
$\sqrt{\frac{1+\sin x}{\cos x}}=\sqrt{\frac{(\cos(x/2)+\sin(x/2))^2}{\cos^2(x/2)-\sin^2(x/2)}}$. After $t=\tan(x/2)$: integral $=\int_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$.
Correct Answer: 2