Question

pls help to solve this indefinate integration question......

Answer 1

@Owlcoffee
@sweetburger @Roxy.girl

Answer 2

@Awolflover1 @AloneS

Answer 3

@braves12

Answer 4

@Mr_Perfection_xD @GIitter @Unavailabilityy @Justsmile_alittle

Answer 5

idk this sorry :/

Answer 6

@ganeshie8

Answer 7

well my answer is in terms of tan inverse not tan

Answer 8

well if u want my solution i can upload it but is not in the options.;/

Answer 9

For a variety, you may differentiate both sides of the given equation

Answer 10

You have \[\int \dfrac{1}{1+\sin x}\,dx = \tan(x/2+a)+b\tag{1}\]

Answer 11

Differentiate both sides with respect to x and get
\[\dfrac{1}{1+\sin x} = \sec^2(x/2+a)*1/2\tag{2}\]

Answer 12

rearrange and try to find the value of \(a\)

Answer 13

\[\dfrac{1}{1+\sin x} = \sec^2(x/2+a)*1/2\]
\[\dfrac{1}{1+\sin x} = \dfrac{1}{2\cos^2(x/2+a)}\]
\[\dfrac{1}{1+\sin x} = \dfrac{1}{1+\cos (x+2a)}\]

Answer 14

That means
\[\sin x = \cos(x+2a)\]

Answer 15

That means
\[\sin x = \cos x\cos(2a) - \sin x\sin(2a)\]

Answer 16

That holds only when \(2a = -\dfrac{\pi}{2}\)

Answer 17

thanks aaaaa loooot... u r definately great.. just like my teacher thanks

Answer 18

np :)