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Evaluate\(\displaystyle\int \dfrac{1}{1+ sinx } dx\)1. tan x + sec x + c2. sec x - tan x + c3. cot x - sec x + c4. tan x - sec x + c5. cot x - tan x + c

Answer» Correct Answer - Option 4 : tan x - sec x + c

Concept:

\(\int sec^2x\ dx = tanx\)

\(\int sec\ x \ tan\ x \ dx= sec\ x\)

Calculation:

\(\displaystyle\int \dfrac{1}{1+ sinx } dx= \displaystyle\int \dfrac{1}{1+ sinx } dx \times \dfrac{1-sinx}{1- sinx}\)

\(=\displaystyle\int \dfrac{1- sinx }{1- sin^2x } dx= \displaystyle\int \dfrac{1- sinx}{cos^2 } dx\)

\(= \displaystyle\int\left[ \dfrac{1}{cos^2x }- \dfrac{sinx}{cos^2 x} dx\right]\)

\(= \displaystyle\int(sec^2 x - sec x. tan x ) dx\)

= tan x - sec x + c



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