The area of the region between the curves y = 1 + sin x cos x and y = 1 - sin x cos x bounded by the lines x = 0 and x = 4 is

Question

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 $x = \frac{π}{4}$ is

Smart Achievers · Accepted Answer

(2)

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

$= \int_{0}^{\frac{π}{4}} \frac{2 \sin \frac{x}{2} d x}{\sqrt{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}}$

$By putting \tan \frac{x}{2} = t, integral will become$

$\int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

Solution

Q.
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 $x = \frac{π}{4}$ is

1 $\int_{\sqrt{2} + 1}^{\sqrt{2} - 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

2 $\int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

3 $\int_{0}^{\sqrt{2} + 1} \frac{4 t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

4 $\int_{0}^{\sqrt{2} + 1} \frac{t}{(1 + t^{2}) \sqrt{1 - t^{2}}} d t$

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Solution

Q. The area of the region between the curves y=1+sinxcosx and y=1-sinxcosx bounded by the lines x=0 and x=π4 is

1 ∫2+12-1t(1+t2)1-t2dt

2 ∫02-14t(1+t2)1-t2dt

3 ∫02+14t(1+t2)1-t2dt

4 ∫02+1t(1+t2)1-t2dt