Sometimes later becomes never. Do it now.

Logic for this lecture:

There are a lot of ways to perform substitution in integration.

First way is to substitute the function only if its derivative is also occurring in question. Sometimes, you need to simplify the function to know whether you can apply substitution or not.

Questions Discussed in this lecture:

21. $$\int \frac{1+\tan x}{1-\tan x} dx = -\log |\cos x-\sin x|+C$$

22. $$\int \frac{\cos 2x}{(\sin x+\cos x)^{2} } dx = \log \left|\sin x+\cos x\right| +C$$

23. $$\int \frac{dx}{x+x\log x} = \log |1+\log x|+C$$

24. $$\int \frac{1}{1+\tan x} dx = \frac{x}{2} +\frac{1}{2} \log |\cos x+\sin x|+C$$

25. $$\int \frac{1}{x-\sqrt{x} } dx = 2\log |\sqrt{x} -1|+C$$

26. $$\int \frac{1}{\cos ^{2} x(1-\tan x)^{2} } dx = \frac{1}{1-\tan x} +C$$

27. $$\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x} dx = \frac{1}{2} \log |2\sin x+3\cos x|+ C$$

28. $$\int \frac{e^{\tan ^{-1} x} }{1+x^{2} } dx = e^{\tan ^{-1} x} +C$$

29. $$\int \frac{e^{2x} -1}{e^{2x} +1} dx = \log |e^{x} +e^{-x} |+ C$$

30. $$\int \sqrt{\sin 2x} \cos 2x dx = \frac{1}{3} (\sin 2x)^{\frac{3}{2} } +C$$