Question
This question will investigate methods for finding definite integrals of powers of trigonometrical functions.
Let $I_n=\int_0^{\frac{\pi}{2}} \sin ^n x d x, n \in \mathrm{N}$.
Let $J_n=\int_0^{\frac{\pi}{2}} \cos ^n x d x, n \in \mathrm{N}$.
Let $T_n=\int_0^{\frac{\pi}{4}} \tan ^n x d x, n \in \mathrm{N}$.
a. Find the exact values of $I_0, I_1$ and $I_2$.
b.i. Use integration by parts to show that $I_n=\frac{n-1}{n} I_{n-2}, n \geqslant 2$.
b.ii.Explain where the condition $n \geqslant 2$ was used in your proof.
c. Hence, find the exact values of $I_3$ and $I_4$.
d. Use the substitution $x=\frac{\pi}{2}-u$ to show that $J_n=I_n$.
e. Hence, find the exact values of $J_5$ and $J_6$
f. Find the exact values of $T_0$ and $T_1$.
g.i. Use the fact that $\tan ^2 x=\sec ^2 x-1$ to show that $T_n=\frac{1}{n-1}-T_{n-2}, n \geqslant 2$.
g.ii.Explain where the condition $n \geqslant 2$ was used in your proof.
h. Hence, find the exact values of $T_2$ and $T_3$.
▶️Answer/Explanation
Markscheme
a.
$$
\begin{aligned}
& I_0=\int_0^{\frac{\pi}{2}} 1 d x=[x]^{\frac{\pi}{2}}=\frac{\pi}{2} \quad \text { M1A1 } \\
& I_1=\int_0^{\frac{\pi}{2}} \sin x d x=[-\cos x]^{\frac{\pi}{2}}=1 \quad \text { M1A1 } \\
& I_2=\int_0^{\frac{\pi}{2}} \sin ^2 x d x=\int_0^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x=\left[\frac{x}{2}-\frac{\sin 2 x}{4}\right]_0^{\frac{\pi}{2}}=\frac{\pi}{4} \quad \text { M1A1 }
\end{aligned}
$$
[6 marks]
b.i. $u=\sin ^{n-1} x$
$v=-\cos x$
$$
\begin{aligned}
& \frac{d u}{d x}=(n-1) \sin ^{n-2} x \cos x \quad \frac{d v}{d x}=\sin x \\
& I_n=\left[-\sin ^{n-1} x \cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}(n-1) \sin ^{n-2} x \cos ^2 x d x \quad \text { M1A1A1 } \\
&=0+\int_0^{\frac{\pi}{2}}(n-1) \sin ^{n-2} x\left(1-\sin ^2 x\right) d x=(n-1)\left(I_{n-2}-I_n\right) \quad \text { M1A1 } \\
& \Rightarrow n I_n=(n-1) I_{n-2} \Rightarrow I_n=\frac{(n-1)}{n} I_{n-2} \quad \text { AG }
\end{aligned}
$$
[6 marks]
b.ii need $n \geqslant 2$ so that $\sin ^{n-1} \frac{\pi}{2}=0$ in $\left[-\sin ^{n-1} x \cos x\right]_0^{\frac{\pi}{2}} \quad \boldsymbol{R} 1$
[1 mark]
C. $I_3=\frac{2}{3} I_1=\frac{2}{3} \quad I_4=\frac{3}{4} I_2=\frac{3 \pi}{16} \quad$ A1A1
[2 marks]
d. $x=\frac{\pi}{2}-u \Rightarrow \frac{d x}{d u}=-1 \quad \boldsymbol{A 1}$
$J_n=\int_0^{\frac{\pi}{2}} \cos ^n x d x=\int_{\frac{\pi}{2}}^0-\cos ^n\left(\frac{\pi}{2}-u\right) d u=-\int_{\frac{\pi}{2}}^0 \sin ^n u d u=\int_0^{\frac{\pi}{2}} \sin ^n u d u=I_n \quad$ M1A1A1AG
[4 marks]
e. $J_5=I_5=\frac{4}{5} I_3=\frac{4}{5} \times \frac{2}{3}=\frac{8}{15} \quad J_6=I_6=\frac{5}{6} I_4=\frac{5}{6} \times \frac{3 \pi}{16}=\frac{5 \pi}{32} \quad$ A1A1
[2 marks]
f.
$$
\begin{aligned}
T_0 & =\int_0^{\frac{\pi}{4}} 1 d x=[x]_0^{\frac{\pi}{4}}=\frac{\pi}{4} \quad \text { A1 } \\
T_1= & \int_0^{\frac{\pi}{4}} \tan d x=[-\ln |\cos x|]_0^{\frac{\pi}{4}}=-\ln \frac{1}{\sqrt{2}}=\ln \sqrt{2} \quad \text { M1A1 }
\end{aligned}
$$
[3 marks]
g.
$$
\begin{aligned}
& \text { g.i. } T_n=\int_0^{\frac{\pi}{4}} \tan ^n x d x=\int_0^{\frac{\pi}{4}} \tan ^{n-2} x \tan ^2 x d x=\int_0^{\frac{\pi}{4}} \tan ^{n-2} x\left(\sec ^2 x-1\right) d x \quad \text { M1 } \\
& \int_0^{\frac{\pi}{4}} \tan ^{n-2} x \sec ^2 x d x-\int_0^{\frac{\pi}{4}} \tan ^{n-2} x d x=\left[\frac{\tan ^{n-1} x}{n-1}\right]_0^{\frac{\pi}{4}}-T_{n-2}=\frac{1}{n-1}-T_{n-2} \quad \text { A1A1AG }
\end{aligned}
$$
[3 marks]
g.ii. need $n \geqslant 2$ so that the powers of $\tan$ in $\int_0^{\frac{\pi}{4}} \tan ^{n-2} x \sec ^2 x d x-\int_0^{\frac{\pi}{4}} \tan ^{n-2} x d x$ are not negative $\quad \mathbf{R}$
[1 mark]
h. $T_2=1-T_0=1-\frac{\pi}{4} \quad$ A1
$$
T_3=\frac{1}{2}-T_1=\frac{1}{2}-\ln \sqrt{2} \quad \text { A1 }
$$
[2 marks]