Question

Do Solve it quickly! ​

Answer 1

Answer:

int(1)/(1+sinx)dx=int(1)/((1+sinx))*((1-sinx))/((1-sinx))dx

=int(1-sinx)/(1-sin^(2)x)dx

=int(1-sinx)/(cos^(2)x)dx

=int(1)/(cos^(2)x)dx-int(sinx)/(cos^(2)x)dx

=int sec^(2)x dx-int tan x sec x dx

=tan x- sec x +C

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Answer 2

Answer:

$\displaystyle \: \int_{-\frac{\pi}{4}}^{ \frac{\pi}{4} } \frac{1}{1 + \sin(x) } dx$

$=\int_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} }( \frac{1}{1 + \sin(x) } \times \frac{1 - \sin(x) }{1 - \sin(x)}dx$

$\displaystyle = \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4} } \frac{ 1 - \sin(x) }{1 - \sin^{2} (x)} dx$

$= \int_{ - \frac{ \pi}{4} }^{ \frac{\pi}{4}} \frac{1}{ \cos {}^{2} (x) } dx - \int_{ - \frac{ \pi}{4} }^{ \frac{\pi}{4} }{ \frac{ \sin(x) }{ \cos {}^{2} (x) } }dx$

$\int_{ - \frac{\pi}{4} }^{ \frac{\pi}{4} } \sec {}^{2} x \: dx - \int_{ - \frac{\pi}{4}}^{ \frac{\pi}{4} } { \tan(x) \sec \: x} \: dx$

$= ((\tan \frac{\pi}{4} - \sec \frac{\pi}{4} )) - (( \tan( - \frac{ \pi}{4} ) - \sec( - \frac{\pi}{4} ))$

$= (1 - \sqrt{2} ) - ( - 1 - \sqrt{2} )$

$= 1 - \cancel{\sqrt{2}} + 1 + \cancel{\sqrt{2} }$

$= 2$