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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 7 Maths Textbook Solution.

Answers (1)

Answer:

x-\tan x+\sec x+c

Given :

\int \frac{\sin x}{1+\sin x} d x

Hint:

Do rationalization and then use trigonometry identities.

Solution:    

\int \frac{\sin x}{1+\sin x} d x

On rationalising,

\int \frac{\operatorname{Sin} x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x

   \int \frac{\operatorname{Sin} x-\sin ^{2}}{1-\sin ^{2} x} d x

  =\int \frac{\sin x}{\cos ^{2} x}-\tan ^{2} \mathrm{x} d x

=\int \frac{\sin x}{\cos x} \times \frac{1}{\cos x}-\left(\sec ^{2} x-1\right) d x

=\int\left(\sec x \tan x-\sec ^{2} x+1\right) d x

=\sec x-\tan x+x+c


 

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