STEP II 2007 question 3 solution

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STEP II 2007 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 3 solution

Part (a)

$x = a\tan(\theta)$

This gives:

$\frac{dx}{d\theta} = a\sec^2(\theta)$

Noting that $1 + \tan^2(\theta) = \sec^2(\theta)$ , substitution gives:

$\int_^\frac{d\theta}{a} = \frac{\theta}{a} + constant$

So the integral is:

$\frac{1}{a}\arctan(\frac{x}{a})$ Q.E.D

Part (b)

$I = \int_0^\frac{\Pi}{2}\frac{\cos x}{1+\sin^2x}dx$

Make the substitution $u = \sin(x)$ which gives:

$\frac{dx}{du} = \frac{1}{\cos x} I = \int_0^1\frac{1}{1+u^2}du = \arctan(1) - \arctan(0) = \frac{\pi}{4}$

Part (c)

As indicated use the following substitution:

$t = \tan(\frac{x}{2})$ giving $\frac{dt}{dx} = 2\sec^2(\frac{x}{2})$

The following identities are used:

$\tan^2(\frac{x}{2}) = \frac{1-\cos x}{1 + \cos x}$ and $\sec^2(\frac{x}{2}) = 1 + \tan^2(\frac{x}{2}) = \frac{2}{1+\cos x}$

The integral becomes:

$\int_0^\frac{\pi}{2}\frac{ \frac{2\cos x}{1+\cos x} }{1 + 6(\frac{1-\cos x}{1+\cos x}) + (\frac{1-\cos x}{1+\cos x})^2 } 0.5\sec^2\frac{x}{2} dx =2\int_0^\frac{\pi}{2}\frac{ \frac{\cos x}{(1+\cos x)^2} }{1 + 6(\frac{1-\cos x}{1+\cos x}) + (\frac{1-\cos x}{1+\cos x})^2 } dx =2\int_0^\frac{\pi}{2}\frac{\cos x}{(1+\cos x)^2+6(1-\cos x)(1+\cos x)+(1-\cos x)^2} dx =2\int_0^\frac{\pi}{2}\frac{\cos x}{2(1+\cos^2 x) + 6\sin^2 x} dx =0.5 \int_0^\frac{\pi}{2}\frac{\cos x}{1+\sin^2 x} dx =0.5 I$

Part (d)

Use the same method as in part (c) to show that this integral becomes:

$0.5\int_0^\frac{\pi}{2}\frac{\cos x}{1+3\sin^2x} dx = 1/6 \int_0^\frac{\pi}{2}\frac{\cos x}{1/3+\sin^2x} dx = 1/6 \int_0^1\frac{1}{1/3 + u^2} du = 1/6 \sqrt{3}\arctan(\sqrt{3}) = 1/6 \sqrt{3} \frac{\pi}{3} = \frac{\pi}{6 \sqrt{3}}$