it's not too hard if you do this sin^3x is sinx(sin^2x)

sin^2x= 1- cos^2x so... integral of (1-cos^2x)sinx dx use substitution now so let u = cosx

du/dx = -sinx so du= -sinxdx take out the -1

-1 integral of (1-u^2)du

-1(u - u^3/3) sub back in the u

-cosx + cos^3x/3 + c

Reply 1390

Pascal678

Original post by R2D2.

for three marks though. And we have limited time... this is messed up

Defo not 3 marks - more like 9 or 10

Reply 1391

Hicko

Original post by Pascal678

Volume of a right angled circular cone =

\dfrac{1}{3} \pi r^{2} h

Volume of a cylinder =

\pi r^{2} h

Area of a circle

\pi r^{2}

Surface area of a cube

6x^{2}

where x is the length of one side of the square i.e. so

x^{2}

is the area of one square

Surface area of a cylinder

2\pi r h +2\pi r^{2}

thank you

Reply 1392

Mutleybm1996

Original post by Pascal678

sin^{3}(x)=sinx(sin^{2}(x))=sinx(1-cos^{2}(x))=sinx-sinxcos^{2}(x)

Could you just post one full solution please? There's so many

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Reply 1393

Vithu09

In this exam, we may forget our maths, we may forget our names, we may forget how to differentiate, hell, we may even forget to do question 9. But we shall never forget to do +C

Reply 1394

Pascal678

Original post by walkers38

.... why ya'll creeping out for sec^3x and e^xsinx are so not gonna come.... i bet my alevel grades on that.

Expect the worse :colone:

Reply 1395

nijam

Original post by Pascal678

Let me quickly show you

\dfrac{du}{dx}=secxtanx

and

v=tanx

so let

I=\displaystyle \int sec^{3}(x) dx = secxtanx-\displaystyle \int secxtan^{2}(x)

Use the fact that

secxtan^{2}(x)=secx(sec^{2}(x)-1)=sec^{3}(x)-secx

Then realise something key here

Unparseable latex formula:

\displaystyle \int sec^{3}(x)=secxtanx-\displaystyle \int sec^{3}(x) dx + \displaystale \int secx dx

This leads to

2 \displaystyle \int sec^{3} (x) = secxtanx+ \displaystyle \int secx dx

Then divide by 2 and integrate the right hand side

is this question in one of the c4 past papers

Reply 1396

ArgieBargy

Original post by Vithu09

In this exam, we may forget our maths, we may forget our names, we may forget how to differentiate, hell, we may even forget to do question 9. But we shall never forget to do +C

On this day, Vithu09 told a big lie, since I always forget +C

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Reply 1397

stephen95

Original post by Mutleybm1996

Could you just post one full solution please? There's so many

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This was my method:

it's not too hard if you do this sin^3x is sinx(sin^2x)

sin^2x= 1- cos^2x so... integral of (1-cos^2x)sinx dx use substitution now so let u = cosx

du/dx = -sinx so du= -sinxdx take out the -1

-1 integral of (1-u^2)du

-1(u - u^3/3) sub back in the u

-cosx + cos^3x/3 + c