What method do you use to calculate $\int \cos^9(x)\sin(x) dx$? (Give me both possibilities!)

A

Substitution ($u = \cos(x)$)

Hint:

Yep - my preferred way to tackle these.

B

Function-derivative

Hint:

Yes, $-\sin(x)$ is the derivative of $\cos(x)$, so you can use function-derivative.

C

Parts ($u = \cos^9(x),~v=\sin(x)$)

Hint:

In principle, that should work... eventually. It's a daft way to do it, though, so I'm marking it wrong.

D

Substitution ($u = \sin(x)$)

Hint:

That's not going to work.

E

Parts ($u = \sin(x),~v' =\cos^9(x)$)

Hint:

I don't know how to integrate $\cos^9(x)$, and neither do you! (Do you?)

Question 5

How do you work out $\int \tan^2(x) dx$?

A

Parts: $u = \tan(x),~v'=\tan(x)$

Hint:

Good luck with integrating $\sec^2(x) \ln(\sec(x)\tan(x))$ in the second step.

B

Parts: $u = \tan^2(x),~v'=1$

Hint:

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.

C

You can only do it numerically.

Hint:

Our survey said: ee-aw.

D

Substitution: start from $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$

Hint:

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

Trig identity: $\tan^2(x) = \sec^2(x) - 1$

Hint:

$sec^2(x)$ integrates to $\tan(x)$. It's in the book.

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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.

Maybe better if you show the answer for the actual maths at the end.

Just telling you but, isn’t the v=1 on the ln(x) question suppose to be v’=1? (2^x has one choice where it is v’=1)
Also, there is 80% chance that “non of the above” is not at the bottom so shouldn’t it be non of the others?

## MathbloggingAll

C4 Integration Quiz (tough stuff) http://t.co/NFQ24eRzbb

## A

Good questions thanks

Maybe better if you show the answer for the actual maths at the end.

Just telling you but, isn’t the v=1 on the ln(x) question suppose to be v’=1? (2^x has one choice where it is v’=1)

Also, there is 80% chance that “non of the above” is not at the bottom so shouldn’t it be non of the others?

## Colin

Thanks for the feedback — I’ve shied away from giving the answers to the integrals, mainly out of laziness.

Thanks especially for the corrections, I’ve fixed them 🙂

## mkami wambura

where are questions?????

## Colin

They should show up after you click the big blue ‘Start’ button.

## aayush ganesh

Differentiating log base 2 is actually fairly straightforward.

y = log2(x)

2^y = x

dx/dy = 2^y ln(2)

dy/dx = 1/((2^y)ln(2))

dy/dx = 1/xln(2)

Imo, this is easier than writing 2^x in terms of e^x and following the e^x integration by recognition rules.

The only reason it’s not used is because it’s not on the syllabus 🙁

They put differentiating and integrating any exponential but not any logarithm. Doesn’t make much sense to me.

## Colin

I’d probably do it as:

$y = \log_2(x) = \frac{\ln(x)}{\ln(2)}$

$\dydx = \frac{1}{x} \times \frac{1}{\ln(2)}$ directly.