Question

Please look at the attachment. Calculus.

Answer 1

We are given with a Indefinite integral , and we need to find it's value ,so , let's start

`{:\implies \quad \displaystyle \sf \int (1)/(1+\sin (x))dx}`

Now , Rationalizing the denominator i.e multiplying the numerator and denominator by the conjugate of denominator i.e 1 - sin(x)

`{:\implies \quad \displaystyle \sf \int \bigg\{(1)/(1+\sin (x))* (1-\sin (x))/(1-\sin (x))\bigg\}dx}`

`{:\implies \quad \displaystyle \sf \int (1-\sin (x))/(1-\sin^(2)(x))dx\quad \qquad \{\because (a-b)(a+b)=a^(2)-b^(2)\}}`

`{:\implies \quad \displaystyle \sf \int (1-\sin (x))/(\cos^(2)(x))dx\quad \qquad \{\because \sin^(2)(x)+\cos^(2)(x)=1\}}`

`{:\implies \quad \displaystyle \sf \int \bigg\{(1)/(\cos^(2)(x))-(\sin (x))/(\cos^(2)(x))\bigg\}dx}`

`{:\implies \quad \displaystyle \sf \int \bigg\{\sec^(2)(x)-(\sin (x))/(\cos (x))* (1)/(\cos (x))\bigg\}dx\quad \qquad \bigg\{\because (1)/(\cos (\theta))=\sec (\theta)\bigg\}}`

`{:\implies \quad \displaystyle \sf \int \{\sec^(2)(x)-\tan (x)\sec (x)\}\quad \qquad \bigg\{\because (\sin (\theta))/(\cos (\theta))=\tan (\theta)\bigg\}}`

Now , we know that ;

• 
  `{\boxed{\displaystyle \bf \int \{f(x)\pm g(x)\}dx=\int f(x)\: dx \pm \int g(x)\: dx}}`

Using this we have ;

`{:\implies \quad \displaystyle \sf \int \sec^(2)(x)dx-\int \tan (x)\sec (x)dx}`

Now , we also knows that ;

• 
  `{\boxed{\displaystyle \bf \int \sec^(2)(x)=\tan (x)+C}}`

• 
  `{\boxed{\displaystyle \bf \int \tan (x)\sec (x)dx=\sec (x)+C}}`

Where C is the Arbitrary Constant . Using this

`{:\implies \quad \displaystyle \sf \tan (x)-\sec (x)+C}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{\displaystyle \bf \int (1)/(1+\sin (x))dx=\tan (x)-\sec (x)+C \:\: \forall \:\: C\in \mathbb{R}}}}`

Answer 2

Multiply the numerator and denominator by 1 - sin(x) :

`(1)/(1 + \sin(x)) * (1 - \sin(x))/(1 - \sin(x)) = (1 - \sin(x))/(1 - \sin^2(x)) = (1-\sin(x))/(\cos^2(x))`

Now separate the terms in the fraction and rewrite them as

`\frac1{\cos^2(x)} - (\sin(x))/(\cos^2(x)) = \sec^2(x) - \tan(x) \sec(x)`

and you'll recognize some known derivatives,

`(d)/(dx) \tan(x) = \sec^2(x)`

`(d)/(dx) \sec(x) = \sec(x) \tan(x)`

So, we have

`\displaystyle \int (dx)/(1 + \sin(x)) = \int (\sec^2(x) - \sec(x) \tan(x)) \, dx = \boxed{\tan(x) - \sec(x) + C}`

which we can put back in terms of sin and cos as

`\tan(x) - \sec(x) = (\sin(x))/(\cos(x))-\frac1{\cos(x)} = (\sin(x)-1)/(\cos(x))`