Question:
$\int\frac{1}{\sin x\, \cos x}$ dx is equal to
$\int\frac{1}{\sin x\, \cos x}$ dx is equal to
Updated On: Jun 6, 2024
- $\log |\tan x| + C$
- $\log |\sin 2x| + C$
- $\log |\sec x| + C$
- $\log |\cos x| + C$
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The Correct Option is A
Solution and Explanation
Let $ I=\int\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right) d x $
$=\int\left(\frac{\sin ^{2} x}{\sin x \cos x}+\frac{\cos ^{2} x}{\sin x \cos x} d x\right)$
$=\int(\tan x+\cot x) d x $
$= \log \sec x+(-\log \operatorname{cosec} x)+C $
$= \log \left|\frac{\sec x}{\operatorname{cosec} x}\right|+C=\log |\tan x|+C $
$=\int\left(\frac{\sin ^{2} x}{\sin x \cos x}+\frac{\cos ^{2} x}{\sin x \cos x} d x\right)$
$=\int(\tan x+\cot x) d x $
$= \log \sec x+(-\log \operatorname{cosec} x)+C $
$= \log \left|\frac{\sec x}{\operatorname{cosec} x}\right|+C=\log |\tan 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.
