Question

Question #5f3d4

Answer 1

`-pi/4`.

Explanation:

We have, `int1/(1+sinx)dx=int1/(1+sinx)xx(1-sinx)/(1-sinx)dx`,

`=int(1-sinx)/cos^2xdx=int(1/cos^2x-sinx/cosx*1/cosx)dx`,

`=int(sec^2x-secxtanx)dx=(tanx-secx)+b`.

But, `tanx-secx=sinx/cosx-1/cosx=(sinx-1)/cosx`,

`=-(1-sinx)/cosx`,

`=-{cos^2(x/2)+sin^2(x/2)-2cos(x/2)sin(x/2)}/(cos^2(x/2)-sin^2(x/2))`,

`=-(cos(x/2)-sin(x/2))^2/{(cos(x/2)-sin(x/2))(cos(x/2)+sin(x/2))`,

`=-(cos(x/2)-sin(x/2))/(cos(x/2)+sin(x/2))`,

`=(sin(x/2)-cos(x/2))/(sin(x/2)+cos(x/2))`,

`={cos(x/2)(sin(x/2)/cos(x/2)-1)}/{cos(x/2)(sin(x/2)/cos(x/2)+1)}`,

`=(sin(x/2)/cos(x/2)-1)/(sin(x/2)/cos(x/2)+1)`,

`=(tan(x/2)-tan(pi/4))/(1+tan(x/2)*tan(pi/4))`.

`rArr int1/(1+sinx)dx=tan(x/2-pi/4)+b`.

`:. int1/(1+sinx)dx=tan(x/2-pi/4)+b=tan(x/2+a)+b`,

`rArr a=-pi/4`, is the desired value!

Answer 2

`-pi/4`.

Explanation:

Here is a Second Method to solve the Problem :

By the Definition of Integral, we have,

`intf(x)dx=F(x)+B iff d/dx{F(x)}=f(x)`.

Accordingly, `int1/(1+sinx)dx=tan(x/2+a)+b`, implies that,

`d/dx{tan(x/2+a)}=1/(1+sinx)`.

`rArr 1/(1+sinx)=sec^2(x/2+a)*d/dx{(x/2+a)}..."[The Chain Rule]"`,

`=1/2*sec^2(x/2+a)=1/(2cos^2(x/2+a))`,

`=1/(1+cos(2*(x/2+a))),.........[because, 1+cos2A=2cos^2A]`,

`=1/(1+cos(x+2a))`,

`=1/[1+sin{pi/2+(x+2a)}]......[because, sin(pi/2+A)=cosA]`,

`i.e., 1/(1+sinx)=1/[1+sin{pi/2+(x+2a)}]`.

`:. pi/2+2a=0, or, a=-pi/4`, as before!

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