# Part - 9 Lecture - 2 Chapter 7 Integrals

Do something today that your future self will thank you for.

Booklets/Notes/Assignments are typed here on this website and their PDFs will be made available soon.

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:

Q68. $$\int_{0}^{\frac{\pi}{2}}sqrt{\sin{\phi}}{\cos}^5{\phi}\phi=\frac{64}{231}$$

Q70.  $$\int_{0}^{1}\left[xe^x+\sin{\frac{\pi x}{4}}\right]dx=1+\frac{4}{\pi}-\frac{2\sqrt2}{\pi}$$

Q71.  $$\int_{0}^{\frac{\pi}{2}}{{\sin}^3{x}dx=\frac{2}{3}}$$

Q72.  $$\int{\frac{1}{sqrt{{\sin}^3{x}{\cos}^5{x}}}dx=-\frac{2}{\sqrt{\tan{x}}}+\frac{2}{3}(\tan{x})^\frac{3}{2}+C}$$

Q73.  $$\int_{0}^{\frac{\pi}{4}}{2{\tan}^3{x}dx}=1-\log{2}$$