We are given with a Indefinite integral , and we need to find it's value ,so , let's start
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{1}{1+\sin (x)}dx}[/tex]
Now , Rationalizing the denominator i.e multiplying the numerator and denominator by the conjugate of denominator i.e 1 - sin(x)
[tex]{:\implies \quad \displaystyle \sf \int \bigg\{\dfrac{1}{1+\sin (x)}\times \dfrac{1-\sin (x)}{1-\sin (x)}\bigg\}dx}[/tex]
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{1-\sin (x)}{1-\sin^{2}(x)}dx\quad \qquad \{\because (a-b)(a+b)=a^{2}-b^{2}\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \int \dfrac{1-\sin (x)}{\cos^{2}(x)}dx\quad \qquad \{\because \sin^{2}(x)+\cos^{2}(x)=1\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \int \bigg\{\dfrac{1}{\cos^{2}(x)}-\dfrac{\sin (x)}{\cos^{2}(x)}\bigg\}dx}[/tex]
[tex]{:\implies \quad \displaystyle \sf \int \bigg\{\sec^{2}(x)-\dfrac{\sin (x)}{\cos (x)}\times \dfrac{1}{\cos (x)}\bigg\}dx\quad \qquad \bigg\{\because \dfrac{1}{\cos (\theta)}=\sec (\theta)\bigg\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \int \{\sec^{2}(x)-\tan (x)\sec (x)\}\quad \qquad \bigg\{\because \dfrac{\sin (\theta)}{\cos (\theta)}=\tan (\theta)\bigg\}}[/tex]
Now , we know that ;
- [tex]{\boxed{\displaystyle \bf \int \{f(x)\pm g(x)\}dx=\int f(x)\: dx \pm \int g(x)\: dx}}[/tex]
Using this we have ;
[tex]{:\implies \quad \displaystyle \sf \int \sec^{2}(x)dx-\int \tan (x)\sec (x)dx}[/tex]
Now , we also knows that ;
- [tex]{\boxed{\displaystyle \bf \int \sec^{2}(x)=\tan (x)+C}}[/tex]
- [tex]{\boxed{\displaystyle \bf \int \tan (x)\sec (x)dx=\sec (x)+C}}[/tex]
Where C is the Arbitrary Constant . Using this
[tex]{:\implies \quad \displaystyle \sf \tan (x)-\sec (x)+C}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\displaystyle \bf \int \dfrac{1}{1+\sin (x)}dx=\tan (x)-\sec (x)+C \:\: \forall \:\: C\in \mathbb{R}}}}[/tex]
