Lethality
Answered

Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Please look at the attachment. Calculus.

Please Look At The Attachment Calculus class=

Sagot :

LammettHash

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

[tex]\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)}[/tex]

Now separate the terms in the fraction and rewrite them as

[tex]\dfrac1{\cos^2(x)} - \dfrac{\sin(x)}{\cos^2(x)} = \sec^2(x) - \tan(x) \sec(x)[/tex]

and you'll recognize some known derivatives,

[tex]\dfrac{d}{dx} \tan(x) = \sec^2(x)[/tex]

[tex]\dfrac{d}{dx} \sec(x) = \sec(x) \tan(x)[/tex]

So, we have

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

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

[tex]\tan(x) - \sec(x) = \dfrac{\sin(x)}{\cos(x)}-\dfrac1{\cos(x)} = \dfrac{\sin(x)-1}{\cos(x)}[/tex]

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]

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.