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: cosx1?
Solution
We start solving the making use of the fact that cosx1=secx. We then multiply and divide the integrand with (secx+tanx) to proceed through the problem. We then assume t=secx+tanx and find dt in terms of dx. We then these results in the integral and then make use of the fact that ∫xdx=lnx+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 cosx1.
Let us assume I=∫cosx1dx ---(1).
We know that cosx1=secx. Let us use this result in equation (1).
⇒I=∫secxdx ---(2).
Let us multiply and divide the integrand with (secx+tanx) in equation (2).
⇒I=∫(secx+tanx)secx(secx+tanx)dx.
⇒I=∫(secx+tanx)(sec2x+secxtanx)dx ---(3).
Let us assume t=secx+tanx ---(4). Now, let us apply a differential on both sides of equation (4).
⇒dt=d(secx+tanx) ---(5).
We know that d(a+b)=da+db. Let us use this result in equation (4).
⇒dt=d(secx)+d(tanx) ---(6).
We know that d(secx)=secxtanxdx and d(tanx)=sec2xdx. Let us use these results in equation (6).
⇒dt=secxtanxdx+sec2xdx.
⇒dt=(secxtanx+sec2x)dx ---(7).
Let us substitute equations (4) and (7) in equation (3).
⇒I=∫tdt ---(8).
We know that ∫xdx=lnx+C. Let us use this result in equation (8).
⇒I=lnt+C.
From equation (3), we have t=secx+tanx.
⇒I=ln(secx+tanx)+C.
∴ We have found the integral of the given function cosx1 as ln(secx+tanx)+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 t with secx+tanx after equation (8), as this is the common mistake done by students. Similarly, we can expect problems to find the integral of the function tanx1.