Question:
The area bounded by the curves $y = cos\, x$ and $y = sin \,x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ is
The area bounded by the curves $y = cos\, x$ and $y = sin \,x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ is
Updated On: Jun 17, 2022
- $\left(4\sqrt{2}-2\right)$ sq units
- $\left(4\sqrt{2}+2\right)$ sq units
- $\left(4\sqrt{2}-1\right)$ sq units
- $\left(4\sqrt{2}+1\right)$ sq units
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The Correct Option is A
Solution and Explanation
Required area
$=\int_\limits{0}^{\pi / 4}(\cos x-\sin x) d x+\int_{\pi / 4}^{5 \pi / 4}(\sin x-\cos x) d x$
$\int_\limits{5 \pi / 4}^{3 \pi / 2}(\cos x-\sin x) d x$
$=[\sin x-\cos x]_{0}^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{5 \pi / 4}$
$+[\sin x-\cos x]_{5 x / 4}^{3 \pi / 4}$
$=(4 \sqrt{2}-2)$ sq unitsa
$=\int_\limits{0}^{\pi / 4}(\cos x-\sin x) d x+\int_{\pi / 4}^{5 \pi / 4}(\sin x-\cos x) d x$
$\int_\limits{5 \pi / 4}^{3 \pi / 2}(\cos x-\sin x) d x$
$=[\sin x-\cos x]_{0}^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{5 \pi / 4}$
$+[\sin x-\cos x]_{5 x / 4}^{3 \pi / 4}$
$=(4 \sqrt{2}-2)$ sq unitsa
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
