Question:
$ \int{\frac{1}{1+\cos \,\,ax}}\,\,dx $ is equal to
$ \int{\frac{1}{1+\cos \,\,ax}}\,\,dx $ is equal to
Updated On: Jun 23, 2024
- $ \cot \,\frac{ax}{2}+c $
- $ \frac{1}{a}\,\tan \,\frac{ax}{2}+c $
- $ \frac{1}{a}(\text{cosec ax - cot ax)+c} $
- $ \frac{1}{a}(\text{cosec ax + cot ax)+c} $
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The Correct Option is B
Solution and Explanation
$ \int{\frac{1}{1+\cos \,ax}}\,dx $
$ =\int{\frac{dx}{2\,{{\cos }^{2}}\,(ax/2)}} $
$ =\frac{1}{2}\int{{{\sec }^{2}}\,\frac{ax}{2}\,dx=\frac{1}{2}}.\frac{\tan ax/2}{a/2} $
$ =\frac{1}{a}\tan \frac{ax}{2}+c $
$ =\int{\frac{dx}{2\,{{\cos }^{2}}\,(ax/2)}} $
$ =\frac{1}{2}\int{{{\sec }^{2}}\,\frac{ax}{2}\,dx=\frac{1}{2}}.\frac{\tan ax/2}{a/2} $
$ =\frac{1}{a}\tan \frac{ax}{2}+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.
