Use integration by parts to find $\int {\left(\text{ln}x\right)}^2\text{d}x$.

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

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

Hence, show that $\int {\text{e}}^x\text{cos}{}^{2}x\text{d}x=\frac{{\text{e}}^x}{5}\text{sin}2x+\frac{{\text{e}}^x}{10}\text{cos}2x+\frac{{\text{e}}^x}{2}+c$,  $c\in \mathbb{R}$.

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

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

Use integration by parts to show that $\int {\text{e}}^x\text{cos}2x\text{d}x=\frac{2{\text{e}}^x}{5}\text{sin}2x+\frac{{\text{e}}^x}{5}\text{cos}2x+c$,  $c\in \mathbb{R}$.

Using the substitution $u=\text{sin}x$, find $\int \frac{\text{co}{\text{s}}^{3}x\text{d}x}{\sqrt{\text{sin}x}}$.

Find $\text{E}\left(X\right)$.

By using the substitution $u=\text{sec} x$ or otherwise, find an expression for $\underset{0}{\overset{\frac{\pi }{3}}{\int }}{\text{sec}}^n x \tan x dx$ in terms of $n$, where $n$ is a non-zero real number.

Hence find the value of $\underset{0}{\overset{1}{\int }}\frac{1}{{\left(x^2+1\right)}^2}\text{d}x$.

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

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

Using the substitution $x=\tan \theta$ show that $\underset{0}{\overset{1}{\int }}\frac{1}{{\left(x^2+1\right)}^2}\text{d}x=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\cos }^2\theta \text{d}\theta$.

By using the substitution $x^2=2\sec \theta$, show that $\int \frac{\text{d}x}{x\sqrt{x^4-4}}=\frac{1}{4}\arccos \left(\frac{2}{x^2}\right)+c$.

Find the area enclosed by the curve and the $x$-axis between B and D, as shaded on the diagram.

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

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

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

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

Find the value of $k$.

Solve the differential equation $\frac{dy}{dx}=\frac{\ln {\displaystyle }{\displaystyle 2}{\displaystyle x}}{x^2}-\frac{2y}{x}, x>0$, given that $y=4$ at $x=\frac{1}{2}$.

Give your answer in the form $y=f\left(x\right)$.

Use the substitution $u=x^{\frac{1}{2}}$ to find $\int \frac{\text{d}x}{x^{\frac{3}{2}}+x^{\frac{1}{2}}}$.

Find the $x$-coordinates of A and of C , giving your answers in the form $a+\text{arctan}b$, where $a$, $b\in \mathbb{R}$.

Find $\int \arcsin x\text{d}x$

By using the substitution $u=\sin x$, find $\int \frac{\sin x \cos x}{{\sin }^2 x-\sin x-2}dx$.

Hence, or otherwise, find $\underset{0}{\overset{\pi }{\int }}{\text{e}}^{-x}\text{sin}x\text{d}x$.