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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: shadezofemerald on September 16, 2008, 07:41:55 pm

Title: Can someone lend a hand with these integration problems?
Post by: shadezofemerald on September 16, 2008, 07:41:55 pm: Please find the integrals of:

1) cosxsin^3x dx

2) sin^2xcos^2x dx

3) sin^3xe^(x/2) dx

any help is greatly appreciated ^_^
Title: Re: Can someone lend a hand with these integration problems?
Post by: Collin Li on September 16, 2008, 08:10:35 pm: Question 1
Use a substitution: $u = \sin x$

Question 2

$\sin^2x\cos^2x = \frac{1}{4}(2\sin x\cos x)^2 = \frac{1}{4}\sin^2(2x)$

Now use $\cos(2A) = 1 - 2\sin^2(A) \implies \sin^2(A) = \frac{1}{2}[1 - \cos(2A)]$ to find:

$\sin^2(2x) = \frac{1}{2}[1 - \cos(4x)]$

Therefore, $\sin^2x\cos^2x = \frac{1}{4}\sin^2(2x) = \frac{1}{8}[1 - \cos(4x)]$

Easy to go from there? ;)

Question 3

$\sin^3x\cdot e^{\frac{x}{2}}$

This isn't required for the Specialist Maths course.
Title: Re: Can someone lend a hand with these integration problems?
Post by: shadezofemerald on September 16, 2008, 08:26:53 pm: thanks coblin for your help ^_^

sorri the third questions was supposed to be: find the integral of sin^3(x/2)
Title: Re: Can someone lend a hand with these integration problems?
Post by: Collin Li on September 16, 2008, 08:30:21 pm: Oh okay, in that case:

Question 3

$\sin^3\left(\frac{x}{2}\right) = \sin\left(\frac{x}{2}\right)\cdot\sin^2\left(\frac{x}{2}\right) = \sin\left(\frac{x}{2}\right) \left[ 1 - \cos^2\left(\frac{x}{2}\right)\right]$

Now, use substitution $u = \cos\left(\frac{x}{2}\right)$ :)
Title: Re: Can someone lend a hand with these integration problems?
Post by: shadezofemerald on September 16, 2008, 08:44:06 pm: ok, thanks heaps!!