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Question: How do you find the integral of \(\left( {\dfrac{{\sec x\tan x}}{{\sec x - 1}}} \right)dx\)?...

How do you find the integral of (sec⁡xtan⁡xsec⁡x−1)dx\left( {\dfrac{{\sec x\tan x}}{{\sec x - 1}}} \right)dx?

Explanation

Solution

According to given in the question we have to find the integral of (sec⁡xtan⁡xsec⁡x−1)dx\left( {\dfrac{{\sec x\tan x}}{{\sec x - 1}}} \right)dx. So, first of all we have to use some formulas to simplify the given expression which are as mentioned below:
Formula used:
⇒sec⁡x=1cos⁡x................(A) ⇒tan⁡x=sin⁡xcos⁡x................(B)  \Rightarrow \sec x = \dfrac{1}{{\cos x}}................(A) \\\ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}................(B) \\\
Now, after substituting the formulas we have to solve the expression obtained by multiplying and dividing the terms of the expression.
Now, we have to let cos⁡x\cos x to some variable which and after that we have to determine the differentiation of the terms we let with respect to x.
Now, we have to factorise the obtained expression with decomposition in simple elements which we have.
Now, we have to solve the integral obtained with the help of the formula to determine the integral which is as explained below:

Formula used:
⇒∫1xdx=log⁡x+c..............(C)\Rightarrow \int {\dfrac{1}{x}dx = \log x + c..............(C)}
Where, c is the constant term.
After that we have to substitute the value of t as we let in the beginning.

Complete step by step solution:
Step 1: First of all we have to use some formulas (A), and (B) to simplify the given expression, which are as mentioned in the solution hint. Hence,
⇒∫1cos⁡x×sin⁡xcos⁡x×11cos⁡x−1dx\Rightarrow \int {\dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{\dfrac{1}{{\cos x}} - 1}}} dx
Step 2: Now, after substituting the formulas we have to solve the expression obtained by multiplying and dividing the terms of the expression. Hence on solving the expression as obtained in the solution step 1,
⇒∫sin⁡xcos⁡x−cos⁡2xdx\Rightarrow \int {\dfrac{{\sin x}}{{\cos x - {{\cos }^2}x}}} dx
Step 3: Now, we have to let cos⁡x\cos x to some variable which and after that we have to determine the differentiation of the terms we let with respect to x. Hence,
⇒t=cos⁡x\Rightarrow t = \cos x
On determining the differentiation of the expression as obtained just above,
⇒dt=−sin⁡x\Rightarrow dt = - \sin x
Step 4: Now, we have to substitute the differentiation as obtained in the solution step 3 in the integral. Hence,
⇒∫1t−t2(−dt)\Rightarrow \int {\dfrac{1}{{t - {t^2}}}( - dt)}
Step 5: Now, we have to factorise the obtained expression with decomposition in simple elements which we have. Hence,
⇒−∫1t(1−t)(dt)\Rightarrow - \int {\dfrac{1}{{t(1 - t)}}(dt)}
Step 6: Now, we have to use the formula (C) which is as mentioned in the solution hint, so that we can determine the integral of the function we obtained in the solution step 5. Hence,
⇒−∫1tdx+∫11−tdt ⇒−[log⁡(t)]+[log⁡(1−t)]  \Rightarrow - \int {\dfrac{1}{t}dx + \int {\dfrac{1}{{1 - t}}dt} } \\\ \Rightarrow - [\log (t)] + [\log (1 - t)] \\\
Step 7: Now, we just have to substitute the value of t as we let in the starting of the solution which is t=cos⁡xt = \cos x.
⇒−[log⁡(cos⁡x)]+[log⁡(1−cos⁡x)]\Rightarrow - [\log (\cos x)] + [\log (1 - \cos x)]
On rearranging the terms of the expression as obtained just above,
⇒2log⁡(sin⁡x2)−log⁡cos⁡x+C\Rightarrow 2\log \left( {\sin \dfrac{x}{2}} \right) - \log \cos x + C

Final solution: Hence, with the help of the formulas (A). (B), and (C) we have determined the integral of(sec⁡xtan⁡xsec⁡x−1)dx\left( {\dfrac{{\sec x\tan x}}{{\sec x - 1}}} \right)dx ⇒2log⁡(sin⁡x2)−log⁡cos⁡x+C \Rightarrow 2\log \left( {\sin \dfrac{x}{2}} \right) - \log \cos x + C.

Note:
It is necessary that we have to let cos⁡x\cos x to some variable and after that we have to substitute the differentiation of the variable we differentiate.
Don’t forget to replace the value of t which we let in the begging of the solution after finding the integration of the function.