Step by step solution to "Solve the integral. ∫cos⁡y1+sin2ydy"

oliviayychengwh

Answered question

2021-12-11

Solve the integral.

\int \frac{\cos y}{\sqrt{1 + \sin^{2} y}} d y

Pademagk71

Beginner2021-12-12Added 34 answers

Step 1
Given,

\int \frac{\cos y}{\sqrt{1 + \sin^{2} y}} d y

Now assuming ,

t = \sin y

then

\frac{d t}{d y} = \cos y

d t = \cos y d y

Sustituting the values of t and dt in the integral

\int \frac{d t}{\sqrt{1 + t^{2}}}

Step 2
Now, we have the formula

\int \frac{d x}{x^{2} + a^{2}} = \ln | x + \sqrt{x^{2} + a^{2}} | + c

Where a is the constant and c is the constant of integration
So,

\int \frac{d t}{\sqrt{1 + t^{2}}} = \ln | t + \sqrt{1 + t^{2}} | + c

Substituting the value of t in the above integral

\int \frac{d t}{\sqrt{1 + t^{2}}} = \ln | \sin y + \sqrt{1 + {(\sin y)}^{2}} | + c

So,

\int \frac{\cos y}{\sqrt{1 + \sin^{2} y}} d y = \ln | \sin y + \sqrt{1 + \sin^{2} y} | + c

(Answer)

Alex Sheppard

Beginner2021-12-13Added 36 answers

\int \frac{\cos (y)}{\sqrt{\sin^{2} (y) + 1}} d y

= \int \frac{1}{\sqrt{u^{2} + 1}} d u

= \int \frac{\sec^{2} (v)}{\sqrt{\tan^{2} (v) + 1}} d v

= \int \sec (v) d v

We expand the fraction by

\tan (v) + \sec (v) :

= \int \frac{\sec (v) (\tan (v) + \sec (v))}{\tan (v) + \sec (v)}

We use the distributive property:

= \int \frac{\sec (v) \tan (v) + \sec^{2} (v)}{\tan (v) + \sec (v)} d v

= \int \frac{1}{w} d w

= \ln (w)

= \ln (\tan (v) + \sec (v))

= \ln (\sqrt{u^{2} + 1} + u)

= \ln (\sqrt{\sin^{2} (y) + 1} + \sin (y))

Solution:

= \ln (\sqrt{\sin^{2} (y) + 1} + \sin (y)) + C

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?