Question

plzz solve yr.. plzz urgent...

Answer 1

Solution:

I have applied decomposition method to find the integral of the given function.

In this method the given non integrable function is modified and splitted into two or more integrable functions.

$\int{\frac{1+sinx}{1-sinx}}\:dx$

$=\int{\frac{1+sinx}{1-sinx}}.\frac{1+sinx}{1+sinx}\:dx$

$=\int{\frac{(1+sinx)^2}{1-sin^{2}x}}\:dx$

$=\int{\frac{1+sin^{2}x+2sinx}{cos^{2}x}}\:dx$

$=\int{[\frac{1}{cos^{2}x}}+\frac{sin^{2}x}{cos^{2}x}+\frac{2sinx}{cos^{2}x}]\:dx$

$=\int{[sec^{2}x+tan^{2}x+2secx\:tanx]\:dx$

$=\int{[sec^{2}x+sec^{2}x-1+2secx\:tanx]\:dx$

$=\int{[2sec^{2}x-1+2secx\:tanx]\:dx$

$=2\int{sec^{2}x}\:dx-\int{dx}+2\int{secx.tanx}\:dx$

$=2tanx-x+2secx+C$

I hope this answer helps you

Answer 2

Answer:

2 tan x + 2 sec x - x + c

Step-by-step explanation:

Given ∫ 1 + sin x/1 – sin x dx

          Multiply and divide by (1 + sin x)

          ∫ (1 + sin x) ^2 / (1 – sin^2 x) dx

         ∫ (1 + sin^2 x + 2 sin x) / cos ^2 x dx

       ∫ (1 / cos ^2 x dx + ∫ 2. Sin x / cos x . 1 / cos x dx + ∫ sin ^2 x / cos ^2 x dx

         ∫ sec ^2 x dx + 2 ∫ tan x. sec x dx + ∫ tan ^2 x dx

         tan x + 2 sec x + tan x + ∫ (sec ^2 x – 1) dx

          tan x + 2 sec x + tan x - x  

            2 tan x + 2 sec x – x + c