2023TrialShort answer

Q12 (b) (iii)

2 marks

Question

Hence or otherwise, evaluate

\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx.

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Solution

(iii)

\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx

=\int_0^{\pi/2}\frac{\frac\pi2-x}{1+\cos\left(\frac\pi2-x\right)+\sin\left(\frac\pi2-x\right)}\,dx\quad\text{from (ii)}

\begin{aligned} \int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx &=\int_0^{\pi/2}\frac{\frac\pi2-x}{1+\sin x+\cos x}\,dx\\ &=\int_0^{\pi/2}\left(\frac{\frac\pi2}{1+\cos x+\sin x}-\frac{x}{1+\cos x+\sin x}\right)dx \end{aligned}

2\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx =\frac\pi2\int_0^{\pi/2}\frac{1}{1+\cos x+\sin x}\,dx

\begin{aligned} &=\frac\pi2(\ln2)\quad\text{from (i)}\\ \therefore\quad \int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx&=\frac\pi4\ln2 \end{aligned}

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Calculus

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

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