Question

How many of the following statements are correct?
$\int tanxdx=ln\left|\right(secx\left)\right|+c\int cotxdx=ln\left|\right(sinx\left)\right|+c\int secxdx=ln\left|\right(tanx\left)\right|+c\int cosec\left(x\right)dx=ln\left|\right(cotx\left)\right|+c$
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Answer 1

All of the above integrals are standard results, which you should be memorizing to solve questions easily. We will find these integrals using substitution.
Let’s find the integral of tanx. For that we will write it as $\frac{-sinx}{cosx}$. We write it this way because we now have a function and it’s derivative. So we can make suitable substitution. After substituting t = cosx, we get the integral equal to - ln |cosx|
$\Rightarrow \int tanxdx=-\int \frac{-sinx}{cosx}dx=-ln\left|cosx\right|+c=ln\left|secx\right|+c$
So the first statement is correct.
Similarly we get the remaining integrals as
$\int cotxdx=\int \frac{cosx}{sinx}dx=ln\left|sinx\right|+c\int secxdx=\int \frac{secx(secx+tanx)}{secx+tanx}dx=ln|secx+tanx|+c$
While finding integral secx , we multiplied and divided by secx+tanx. This is to make the numerator, derivative of denominator.
We do the same for cosecx also, multiply by cosecx - cotx
$\int cosecxdx=\int \frac{cosecx(cotx-cosecx)}{cotx-cosecx}dx=ln(cotx-cosecx)+c$
So, only first two statements are correct
We can verify all these results by taking derivative of the RHS and checking if we get the function we are trying to integrate.