Find the exact values of $I_0$, $I_1$ and $I_2$.

Use integration by parts to show that $I_n=\frac{n-1}{n}{I}_{n-2}\text{,}n⩾2$.

Explain where the condition $n⩾2$ was used in your proof.

Hence, find the exact values of $I_3$ and $I_4$.

Use the substitution $x=\frac{\pi }{2}-u$ to show that $J_n=I_n$.

Hence, find the exact values of $J_5$ and $J_6$

Find the exact values of $T_0$ and $T_1$.

Use the fact that $\text{ta}{\text{n}}^{2}x=\text{se}{\text{c}}^{2}x-1$ to show that $T_n=\frac{1}{n-1}-T_{n-2}\text{,}n⩾2$.

Explain where the condition $n⩾2$ was used in your proof.

Hence, find the exact values of $T_2$ and $T_3$.

$I_0=\underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx={\left[x\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$      M1A1

$I_1=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\text{sin}xdx={\left[-\text{cos}x\right]}_0^{\frac{\pi }{2}}=1$      M1A1

$I_2=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\text{si}{\text{n}}^{2}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-\text{cos}2x}{2}dx={\left[\frac{x}{2}-\frac{\text{sin}2x}{4}\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{4}$      M1A1

[6 marks]

$u=\text{si}{\text{n}}^{n-1}x$                               $v=-\cos x$

$\frac{du}{dx}=\left(n-1\right)\text{si}{\text{n}}^{n-2}x\text{cos}x$     $\frac{dv}{dx}=\text{sin}x$

$I_n={\left[-\text{si}{\text{n}}^{n-1}x\text{cos}x\right]}_0^{\frac{\pi }{2}}+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(n-1\right)\text{si}{\text{n}}^{n-2}x\text{cos}{}^{2}xdx$      M1A1A1

$=0+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(n-1\right)\text{si}{\text{n}}^{n-2}x\left(1-\text{si}{\text{n}}^{2}x\right)dx=\left(n-1\right)\left(I_{n-2}-I_n\right)$      M1A1

$\Rightarrow n{I}_n=\left(n-1\right)I_{n-2}\Rightarrow I_n=\frac{\left(n-1\right)}{n}{I}_{n-2}$        AG

[6 marks]

need $n⩾2$ so that $\text{si}{\text{n}}^{n-1}\frac{\pi }{2}=0$ in ${\left[-\text{si}{\text{n}}^{n-1}x\text{cos}x\right]}_0^{\frac{\pi }{2}}$       R1

[1 mark]

$I_3=\frac{2}{3}{I}_1=\frac{2}{3}{I}_4=\frac{3}{4}{I}_2=\frac{3\pi }{16}$      A1A1

[2 marks]

$x=\frac{\pi }{2}-u\Rightarrow \frac{dx}{du}=-1$      A1

$J_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\text{co}{\text{s}}^{n}xdx=\underset{\frac{\pi }{2}}{\overset{0}{\int }}-\text{co}{\text{s}}^n\left(\frac{\pi }{2}-u\right)du=-\underset{\frac{\pi }{2}}{\overset{0}{\int }}\text{si}{\text{n}}^{n}udu=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\text{si}{\text{n}}^{n}udu=I_n$      M1A1A1AG

[4 marks]

$J_5=I_5=\frac{4}{5}{I}_3=\frac{4}{5}\times \frac{2}{3}=\frac{8}{15}{J}_6=I_6=\frac{5}{6}{I}_4=\frac{5}{6}\times \frac{3\pi }{16}=\frac{5\pi }{32}$     A1A1

[2 marks]

$T_0\text{ = }\underset{0}{\overset{\frac{\pi }{4}}{\int }}1dx={\left[x\right]}_0^{\frac{\pi }{4}}=\frac{\pi }{4}$      A1

$T_1\text{ = }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{tan}dx={\left[-\ln \left|\text{cos}x\right|\right]}_0^{\frac{\pi }{4}}=-\text{ln}\frac{1}{\sqrt{2}}=\text{ln}\sqrt{2}$       M1A1

[3 marks]

$T_n=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n}xdx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}x\text{ta}{\text{n}}^{2}xdx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}x\left(\text{se}{\text{c}}^{2}x-1\right)dx$         M1

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}x\text{se}{\text{c}}^{2}xdx-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}xdx={\left[\frac{\text{ta}{\text{n}}^{n-1}x}{n-1}\right]}_0^{\frac{\pi }{4}}-T_{n-2}=\frac{1}{n-1}-T_{n-2}$        A1A1AG

[3 marks]

need $n⩾2$ so that the powers of tan in $\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}x\text{se}{\text{c}}^{2}xdx-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\text{ta}{\text{n}}^{n-2}xdx$ are not negative         R1   

[1 mark]

$T_2=1-T_0=1-\frac{\pi }{4}$         A1 

$T_3=\frac{1}{2}-T_1=\frac{1}{2}-\ln \sqrt{2}$         A1

[2 marks]