Evaluate the integral. ∫cos⁡x2+sin⁡x

Question

Evaluate the integral.  ∫cos⁡x2+sin⁡x

Robert Pina · Accepted Answer

Step 1
Our Aim is to evaluate the integral given below:

\int \frac{\cos (x)}{2 + \sin (x)} d x

...(i)
Step 2
Considering the integral given by equation−(i), we have:−

\int \frac{\cos (x)}{2 + \sin (x)} d x

For the integrand

\frac{\cos (x)}{2 + \sin (x)}

, we will substitute

u = \sin (x) + 2

and

d u = \cos (x) d x

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

Since, the integral of

\frac{1}{u}

is

\log (u)

= \log (u)

+constant.
Substituting back for

u = \sin (x) + 2

Answer

\Rightarrow \log (\sin (x) + 2)

+ constant.

Karen Robbins · Accepted Answer

Given that ∫cos⁡x2+sin⁡xdx  Let 2+sin⁡x=u  ⇒cos⁡xdx=du  Now ∫cos⁡x2+sin⁡xdx=∫1udu  =ln|u|+c  ∴∫cos⁡x2+sin⁡xdx=ln|2+sin⁡x|+c

Vasquez · Accepted Answer

∫cos⁡x2+sin⁡xdx We put the expression cos⁡(x) under the differential sign, i.e.: cos⁡(x)dx=d(sin⁡(x)),t=sin⁡(x) Then the original integral can be written as follows: ∫1t+2dt ∫1x+2dx Calculate the table integral: ∫1x+2dx=ln⁡(x+2) Answer: ln⁡(x+2)+c To write the final answer, it remains to substitute sin⁡(x) instead of t. ln⁡(sin⁡(x)+2)+C

Evaluate the integral. \int \frac{\cos x}{2+\sin x}

Answered question

Answer & Explanation

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