Question

Please look at the attachment. Calculus.

Answer 1

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

$\dfrac{1}{1 + \sin(x)} \times \dfrac{1 - \sin(x)}{1 - \sin(x)} = \dfrac{1 - \sin(x)}{1 - \sin^2(x)} = \dfrac{1-\sin(x)}{\cos^2(x)}$

Now separate the terms in the fraction and rewrite them as

$\dfrac1{\cos^2(x)} - \dfrac{\sin(x)}{\cos^2(x)} = \sec^2(x) - \tan(x) \sec(x)$

and you'll recognize some known derivatives,

$\dfrac{d}{dx} \tan(x) = \sec^2(x)$

$\dfrac{d}{dx} \sec(x) = \sec(x) \tan(x)$

So, we have

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

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

$\tan(x) - \sec(x) = \dfrac{\sin(x)}{\cos(x)}-\dfrac1{\cos(x)} = \dfrac{\sin(x)-1}{\cos(x)}$

Answer 2

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

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

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

${:\implies \quad \displaystyle \sf \int \bigg\{\dfrac{1}{1+\sin (x)}\times \dfrac{1-\sin (x)}{1-\sin (x)}\bigg\}dx}$

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

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

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

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

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

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

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

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