Question:
$\int\frac{2x+\sin2x}{1+\cos2x}\, dx$ is equal to
$\int\frac{2x+\sin2x}{1+\cos2x}\, dx$ is equal to
Updated On: Apr 8, 2024
- $x +\log |\tan x| + C$
- $x \log |\tan x| + C$
- $x \tan x + C$
- $\log |\cos x| + C$
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The Correct Option is C
Solution and Explanation
Let $ I=\int \frac{2 x}{1+\cos 2 x} $
$=\int \frac{2 x}{1+\cos 2 x}+\frac{\sin 2 x}{1+\cos 2 x} d x $
$=\int \frac{2 x}{2 \cos ^{2} x}+\int \tan x d x $
$=\int x \sec ^{2} x+\int \tan x d x $
$=x \tan x-\int \tan x d x+\int \tan x d x $
$=x \tan x+C $
$=\int \frac{2 x}{1+\cos 2 x}+\frac{\sin 2 x}{1+\cos 2 x} d x $
$=\int \frac{2 x}{2 \cos ^{2} x}+\int \tan x d x $
$=\int x \sec ^{2} x+\int \tan x d x $
$=x \tan x-\int \tan x d x+\int \tan x d x $
$=x \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.
