2022TrialShort answer

Q12 (b)

3 marks

Question

Evaluate
$\int_0^{\pi/4}\sec^4x\tan^2x\,dx.$

Try the question before revealing the answer.

Solution

\int_0^{\pi/4}\sec^4x\tan^2x\,dx =\int_0^{\pi/4}\sec^2x\sec^2x\tan^2x\,dx

=\int_0^{\pi/4}\sec^2x(1+\tan^2x)\tan^2x\,dx

=\int_0^{\pi/4}\left(\sec^2x\tan^2x+\sec^2x\tan^4x\right)dx

=\left[\frac{\tan^3x}{3}+\frac{\tan^5x}{5}\right]_0^{\pi/4}

=\left(\frac{\tan^3(\pi/4)}{3}+\frac{\tan^5(\pi/4)}{5}\right)-\left(\frac{\tan^3(0)}{3}+\frac{\tan^5(0)}{5}\right)

=\frac13+\frac15=\frac{8}{15}

Alternative Solution

\int_0^{\pi/4}\sec^4x\tan^2x\,dx =\int_0^{\pi/4}4(1+\tan^2x)\tan^2x\sec^2x\,dx

Let $u=\tan x$ .

When $x=\frac{\pi}{4}$ , $u=1$ ; when $x=0$ , $u=0$ .

du=\sec^2x\,dx

\int_0^1(1+u^2)u^2\,du =\left[\frac{u^3}{3}+\frac{u^5}{5}\right]_0^1

=\frac{8}{15}

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Syllabus alignment

Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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