University Maths Solution
| Maths Question | |
|---|---|
| Question 41 |
$\int{{{\tan }^{3}}x{{\sec }^{2}}xdx}$ |
| Question 42 |
$\int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}$ |
| Question 43 |
$\begin{align} & \text{Show that }\frac{d}{dx}\left( \frac{1}{\sin x\cos x} \right)=\frac{2}{{{\cos }^{2}}x}-\frac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}, \\ & \text{Hence or otherwise evaluate}\int{\frac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}} \\\end{align}$ |
| Question 44 |
$\text{Evaluate}\int{x\sqrt{x+1}dx}$ |
| Question 45 |
$\text{Evaluate }\int{\frac{{{x}^{4}}}{{{({{x}^{5}}+6)}^{2}}}dx}$ |
| Question 46 |
$\begin{align} & \text{Use substitution }x=\cos 2\theta ,\text{ to prove that} \\ & \text{ }\int_{-1}^{1}{\sqrt{\frac{1+x}{1-x}}dx}=\pi \\\end{align}$ |
| Question 47 |
$\text{Prove that }\int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}\sin x\cos xdx}={{\pi }^{2}}-\frac{1}{4}$ |
| Question 48 |
$\text{Evaluate }\int_{0}^{\tfrac{\pi }{4}}{(2x{{\sec }^{2}}{{x}^{2}}+\tan x)dx}$ |
| Question 49 |
$\text{Show that }\int_{1}^{2}{x\log ({{x}^{2}}+1)dx}=\frac{5}{2}{{\log }_{e}}5-{{\log }_{e}}2-\frac{3}{2}$ |
| Question 50 |
$\text{Show that }\int_{\tfrac{\pi }{3}}^{\tfrac{\pi }{2}}{\frac{\cos xdx}{1-\cos x}}$ |
| Question 51 |
$\text{By means of substitution of }x=\frac{1}{u}\text{ or otherwise, evaluate }\int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$ |
| Question 52 |
$\text{Evaluate }\int{\frac{\sin 2\theta }{\sec 2\theta }d\theta }$ |
| Question 53 |
$\int_{0}^{\tfrac{\pi }{2}}{(\sin 2\theta -\sin \theta )d\theta }$ |