Here’s a quick multiple-choice quiz about the tough stuff in C4 integration.

Ready? ((If you’re not, you should buy my book on C4 integration.))

### Question 1: squared trig functions

What method do you use to calculate $\int \sin^2(x) dx$? (Give me all four answers!)

a) Parts ($u = \sin(x),~v’=\sin(x)$) b) Trig substitution ($u=\cos(2x)$) c) Split-angle formula ($\sin(A)\sin(B) = d) Parts ($u = \sin^2(x),~v’=1$) e) None of the above. ### Question 2: mixed trig functions What method do you use to calculate$\int \cos^9(x)\sin(x) dx$? (Give me both possibilities!) a) Substitution ($u = \sin(x)$) b) Function-derivative c) Parts ($u = \sin(x),~v’ =\cos^9(x)$) d) Parts ($u = \cos^9(x),~v=\sin(x)$) e) Substitution ($u = \cos(x)$) ### Question 3: logarithms How do you work out$\int \ln(x) dx$? (I want two methods.) a) Parts ($u = \ln(x),~v=1$) b) You look it up - it’s$\frac{1}{x} + C$c) Parts ($u = 1,~v’=\ln(x)$) d) Substitution:$x = e^u$e) You can only do it numerically ### Question 4: Nasty powers What method do you use to calculate$\int 2^x dx$? a) Parts:$u = 2^x,~v’=1$b) Substitute$u = log_2(x)$c) Increase the power and divide by the new power. d) Replace$2^x$with$e^{x\ln(2)}$. e) You can only do it numerically ### Question 5: Another squared trig function How do you work out$\int \tan^2(x) dx$? a) Trig identity:$\tan^2(x) = \sec^2(x) - 1$b) Parts:$u = \tan(x),~v’=\tan(x)$c) Parts:$u = \tan^2(x),~v’=1$d) Substitution: start from$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$e) You can only do it numerically. ## Answers Let’s go through them and see what our hundred people said ((Just kidding. No people were harmed in the making of this quiz.)) ### Question 1: squared trig functions What method do you use to calculate$\int \sin^2(x) dx$? (Give me all four answers!) a) Parts ($u = \sin(x),~v’=\sin(x)$) ✓ You can do! You need to use a trig identity in the next step, though. b) Trig substitution ($u=\cos(2x)$) ✓ Probably the easiest way. c) Split-angle formula ✓ The way you get most help from the book with d) Parts ($u = \sin^2(x),~v’=1$) ✓ Probably the most involved way - you have to do parts and a trig substitution in the second step. e) None of the above ✕ Nope - the clue’s in the question! ### Question 2: mixed trig functions What method do you use to calculate$\int \cos^9(x)\sin(x) dx$? (Give me both possibilities!) a) Substitution ($u = \sin(x)$) ✕ That’s not going to work - or at least, not easily. b) Function-derivative ✓ Yes,$-\sin(x)$is the derivative of$\cos(x)$, so you can use function-derivative. c) Parts ($u = \sin(x),~v’ =\cos^9(x)$) ✕ I don’t know how to integrate$\cos^9(x)$, and neither do you! (Do you?) d) Parts ($u = \cos^9(x),~v=\sin(x)$) ½ In principle, that should work… eventually. It’s a daft way to do it, though. e) Substitution ($u = \cos(x)$) ✓ Yep - my preferred way to tackle these. ### Question 3: logarithms How do you work out$\int \ln(x) dx$? (I want two methods.) a) Parts ($u = \ln(x),~v=1$) ✓ Yep, it drops out nicely. b) You look it up - it’s$\frac{1}{x} + C$✕ No, that’s differentiating c) Parts ($u = 1,~v’=\ln(x)$) ✕ No, if you knew how to integrate$ln$, you wouldn’t be in this mess. d) Substitution:$x = e^u$✓ Not a popular way, but a good way. e) You can only do it numerically ✕ Nut-uh. You _can do it numerically, of course, but it’s not the only way._ ### Question 4: Nasty powers What method do you use to calculate$\int 2^x dx$? a) Parts:$u = 2^x,~v’=1$✕ Good god, no. Have an ibuprofen. b) Substitute$u = log_2(x)$½ … sorta. Differentiating$log_2(x)$isn’t trivial, though. c) Increase the power and divide by the new power. ✕ ✕ ✕ YOU KILLED A KITTEN, YOU BASTARD! d) Replace$2^x$with$e^{x\ln(2)}$. ✓ Yep - messy, but it works. e) You can only do it numerically ✕ That’s not true. ### Question 5: Another squared trig function How do you work out$\int \tan^2(x) dx$? a) Trig identity:$\tan^2(x) = \sec^2(x) - 1$: ✓ Yep,$sec^2(x)$integrates to$\tan(x)$. It’s in the book. b) Parts:$u = \tan(x),~v’=\tan(x)$: ✕ Good luck with integrating$\sec^2(x) \ln(\sec(x)\tan(x))$in the second step. c) Parts:$u = \tan^2(x),~v’=1$✕ This ends up as$x\tan^2(x) - \int 2x \tan^2(x)sec(x) dx$. I reckon it’s possible, but I don’t fancy it. d) Substitution: start from$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$✕ It’s a nice thought, but no. You end up with a$\frac{\tan(2x)}{\tan(x)}\$ to integrate, which is no good at all.

e) You can only do it numerically. ✕ Who gave you that idea?