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Question

Question: How do you find the integral of the given function: \(\dfrac{1}{\cos x}\)?...

How do you find the integral of the given function: 1cosx\dfrac{1}{\cos x}?

Explanation

Solution

We start solving the making use of the fact that 1cosx=secx\dfrac{1}{\cos x}=\sec x. We then multiply and divide the integrand with (secx+tanx)\left( \sec x+\tan x \right) to proceed through the problem. We then assume t=secx+tanxt=\sec x+\tan x and find dtdt in terms of dxdx. We then these results in the integral and then make use of the fact that dxx=lnx+C\int{\dfrac{dx}{x}}=\ln x+C and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the integral of the given function 1cosx\dfrac{1}{\cos x}.
Let us assume I=1cosxdxI=\int{\dfrac{1}{\cos x}dx} ---(1).
We know that 1cosx=secx\dfrac{1}{\cos x}=\sec x. Let us use this result in equation (1).
I=secxdx\Rightarrow I=\int{\sec xdx} ---(2).
Let us multiply and divide the integrand with (secx+tanx)\left( \sec x+\tan x \right) in equation (2).
I=secx(secx+tanx)(secx+tanx)dx\Rightarrow I=\int{\dfrac{\sec x\left( \sec x+\tan x \right)}{\left( \sec x+\tan x \right)}dx}.
I=(sec2x+secxtanx)(secx+tanx)dx\Rightarrow I=\int{\dfrac{\left( {{\sec }^{2}}x+\sec x\tan x \right)}{\left( \sec x+\tan x \right)}dx} ---(3).
Let us assume t=secx+tanxt=\sec x+\tan x ---(4). Now, let us apply a differential on both sides of equation (4).
dt=d(secx+tanx)\Rightarrow dt=d\left( \sec x+\tan x \right) ---(5).
We know that d(a+b)=da+dbd\left( a+b \right)=da+db. Let us use this result in equation (4).
dt=d(secx)+d(tanx)\Rightarrow dt=d\left( \sec x \right)+d\left( \tan x \right) ---(6).
We know that d(secx)=secxtanxdxd\left( \sec x \right)=\sec x\tan xdx and d(tanx)=sec2xdxd\left( \tan x \right)={{\sec }^{2}}xdx. Let us use these results in equation (6).
dt=secxtanxdx+sec2xdx\Rightarrow dt=\sec x\tan xdx+{{\sec }^{2}}xdx.
dt=(secxtanx+sec2x)dx\Rightarrow dt=\left( \sec x\tan x+{{\sec }^{2}}x \right)dx ---(7).
Let us substitute equations (4) and (7) in equation (3).
I=dtt\Rightarrow I=\int{\dfrac{dt}{t}} ---(8).
We know that dxx=lnx+C\int{\dfrac{dx}{x}}=\ln x+C. Let us use this result in equation (8).
I=lnt+C\Rightarrow I=\ln t+C.
From equation (3), we have t=secx+tanxt=\sec x+\tan x.
I=ln(secx+tanx)+C\Rightarrow I=\ln \left( \sec x+\tan x \right)+C.

\therefore We have found the integral of the given function 1cosx\dfrac{1}{\cos x} as ln(secx+tanx)+C\ln \left( \sec x+\tan x \right)+C.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to add constant of integration while solving the problems related to indefinite integrals. We should not forget to replace tt with secx+tanx\sec x+\tan x after equation (8), as this is the common mistake done by students. Similarly, we can expect problems to find the integral of the function 1tanx\dfrac{1}{\tan x}.