Question:
$ \int{\frac{\cos x-\sin x}{1+2\sin x\cos x}}dx $ is equal to
$ \int{\frac{\cos x-\sin x}{1+2\sin x\cos x}}dx $ is equal to
Updated On: Jun 6, 2022
- $ -\frac{1}{\cos x-\sin x}+C $
- $ \frac{\cos x+\sin x}{\cos x-\sin x}+C $
- $ -\frac{1}{\sin x+\cos x}+C $
- $ \frac{1}{\sin x+\cos x}+C $
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The Correct Option is C
Solution and Explanation
Let $ I=\int{\frac{\cos x-\sin x}{{{(\cos x+\sin x)}^{2}}}}dx $
Put $ \cos x+\sin x=t $
$ \Rightarrow $ $ (\cos x-\sin x)dx=dt $
$ \therefore $ $ I=\int{\frac{1}{{{t}^{2}}}}dt $
$ I=-\frac{1}{t}+c=-\frac{1}{\sin x+\cos x}+c $
Put $ \cos x+\sin x=t $
$ \Rightarrow $ $ (\cos x-\sin x)dx=dt $
$ \therefore $ $ I=\int{\frac{1}{{{t}^{2}}}}dt $
$ I=-\frac{1}{t}+c=-\frac{1}{\sin x+\cos x}+c $
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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.
