Evaluate the integral. ∫1−sin⁡xcos⁡xdx

Question

Evaluate the integral.  ∫1−sin⁡xcos⁡xdx

Jim Hunt · Accepted Answer

Step 1
We have to evaluate the integral:

\int \frac{1 - \sin x}{\cos x} d x

We will use following special integration formula:

\int \sec x d x = \log | \sec x + \tan x | + c

\int \tan x d x = \log | \sec x | + c

Step 2
So simplifying further,

\int \frac{1 - \sin x}{\cos x} d x = \int (\frac{1}{\cos x} - \frac{\sin x}{\cos x}) d x

= \int (\sec x - \tan x) d x

= \int \sec x d x - \int \tan x d x

= \log | \sec x + \tan x | - \log | \sec x | + c

= \log (| \frac{\sec x + \tan x}{\sec x} |) + c

= \log (| \frac{\sec x}{\sec x} + \frac{\tan x}{\sec x} |) + c

= \log (| 1 + \cos x |) + c

Hence, value of integral is

\log (| 1 + \cos x |)

.

Bubich13 · Accepted Answer

∫1−sin⁡(x)cos⁡(x)dx  =∫(1cos⁡x−sin⁡(x)cos⁡(x))dx  =∫1cos⁡(x)dx−∫sin⁡(x)cos⁡(x)dx  ∫1cos⁡(x)dx  =∫sec⁡(x)dx  =ln⁡(tan⁡(x)+sec⁡(x))  ∫sin⁡(x)cos⁡(x)dx  =−∫1udu  ∫1udu  =ln⁡(u)  −∫1udu  =−ln⁡(u)  =−ln⁡(cos⁡(x))  ∫1cos⁡(x)dx−∫sin⁡(x)cos⁡(x)dx  =ln⁡(tan⁡(x)+sec⁡(x))+ln⁡(cos⁡(x))  Answer:  =ln⁡(|tan⁡(x)+sec⁡(x)|)+ln⁡(|cos⁡(x)|)+C

Evaluate the integral. \int \frac{1-\sin x}{\cos x}dx

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