What is / x+ x equal to?

Correct Answer is : ln | sec x + tan x | - ln | sec x | + c

ln (sec x) + ln | sec x + tan x | + c

ln (sec x) - ln | sec x + tan x | + c

sec x tan x - ln | sec x - tan x | + c

Correct Answer is : ln | sec x + tan x | - ln | sec x | + c

Test Index

UPSC NDA I 2021 Mathematics Solved Paper

Question : 100 of 120

Marks: +1, -0

What is

\int\frac{dx}{\sec x+\tan x}

equal to?

Solution:

Concept:

Use the identity

1 + \tan^2 x = \sec^2 x

, which gives

\sec^2 x - \tan^2 x = 1

Standard integrals:

\int \sec x \, dx = \ln |\sec x + \tan x| + C

and

\int \tan x \, dx = \ln |\sec x| + C

Explanation:

Let

I = \int \frac{dx}{\sec x + \tan x}

Rationalize the denominator by multiplying numerator and denominator by

(\sec x - \tan x)

I = \int \frac{(\sec x - \tan x)}{(\sec x + \tan x)(\sec x - \tan x)} \, dx = \int \frac{\sec x - \tan x}{\sec^2 x - \tan^2 x} \, dx

Since

\sec^2 x - \tan^2 x = 1

, the integral simplifies to

I = \int (\sec x - \tan x) \, dx

Split the integral:

\int \sec x \, dx - \int \tan x \, dx

Apply the standard integrals:

\int \sec x \, dx = \ln |\sec x + \tan x|

and

\int \tan x \, dx = \ln |\sec x|

Combining constants of integration, we get

I = \ln |\sec x + \tan x| - \ln |\sec x| + C

Answer:

\ln |\sec x + \tan x| - \ln |\sec x| + C

(Option D).

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