# C4 Integration Quiz (tough stuff)

Here's a quick multiple-choice quiz about the tough stuff in 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 said2 ### 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? ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. If you're not, you should buy my book on C4 integration. []
2. Just kidding. No people were harmed in the making of this quiz. []

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