What is dx/ x + x equal to? [NDA-I 2021]

Correct Answer is : ln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+c | x+ x|- | x|+cln∣secx+tanx∣−ln∣secx∣+c

Correct Answer is : ln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+c | x+ x|- | x|+cln∣secx+tanx∣−ln∣secx∣+c

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UPSC NDA I 2021 Mathematics Solved Paper

Question : 100 of 120

Marks: +1, -0

What is

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

Solution: 👈: Video Solution

Concept:

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

. -

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

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

Calculation:

Let

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

By rationalizing the denominator of the integrand we get:

\Rightarrow I = \int \frac{(\sec x - tanx)}{(\sec x + \tan x) \cdot (\sec x - \tan x)} dx

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

As we know that,

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

therefore:

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

\Rightarrow I = \int \sec x dx - \int \tan x dx

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

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