10 questions yellow

We are given:

$\int_0^1 \dfrac{x^2}{(1 + x^2)^3} \, dx$Use the substitution $x = \tan u$.

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Step 1: Substitution and Limits

Let $x = \tan u$, so $dx = \sec^2 u , du$.
Also, since $x = 0 \Rightarrow u = 0$ and $x = 1 \Rightarrow u = \dfrac{\pi}{4}$.

Now, substitute into the integral:

$$\begin{aligned} \int_0^1 \dfrac{x^2}{(1 + x^2)^3} \, dx &= \int_0^{\frac{\pi}{4}} \dfrac{\tan^2 u}{(1 + \tan^2 u)^3} \cdot \sec^2 u \, du \end{aligned}$$ 

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Step 2: Simplify the Expression

Recall the identity $1 + \tan^2 u = \sec^2 u$:

$$\begin{array}{rl}{\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2}u}{{\sec }^{6}u}\cdot {\sec }^{2}udu}& {\displaystyle ={\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2}u}{{\sec }^{4}u}du}\\ {\displaystyle }& {\displaystyle ={\int }_0^{\frac{\pi }{4}}{\tan }^{2}u\cdot {\cos }^{4}udu}\end{array}$$Because $\dfrac{1}{\sec^4 u} = \cos^4 u$ and $\tan u = \dfrac{\sin u}{\cos u}$:

$$\tan^2 u \cdot \cos^4 u = \left( \dfrac{\sin^2 u}{\cos^2 u} \right) \cdot \cos^4 u = \sin^2 u \cdot \cos^2 u$$ 

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Step 3: Final Integral

Now the integral becomes:

$${\int }_0^{\frac{\pi }{4}}{\sin }^{2}u\cdot {\cos }^{2}udu$$Use the identity:
$\sin^2 u \cdot \cos^2 u = \dfrac{1}{4} \sin^2 2u$
(because $\sin 2u = 2 \sin u \cos u$)

So:

$${\int }_0^{\frac{\pi }{4}}\frac{1}{4}{\sin }^{2}2udu$$Use the identity:
$\sin^2 2u = \dfrac{1 – \cos 4u}{2}$

Then:

$$\int_0^{\frac{\pi}{4}} \dfrac{1}{4} \cdot \dfrac{1 – \cos 4u}{2} \, du = \int_0^{\frac{\pi}{4}} \dfrac{1 – \cos 4u}{8} \, du$$ 

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Step 4: Integrate

Now compute:

$${\int }_0^{\frac{\pi }{4}}\frac{1-\cos 4u}{8}du=\frac{1}{8}{\int }_0^{\frac{\pi }{4}}(1-\cos 4u)du$$Break it up:

$$= \dfrac{1}{8} \left[ u – \dfrac{\sin 4u}{4} \right]_0^{\frac{\pi}{4}} = \dfrac{1}{8} \left[ \dfrac{\pi}{4} – \dfrac{\sin \pi}{4} \right] = \dfrac{1}{8} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{32}$$ 

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Final Answer

$\boxed{{\displaystyle \frac{\pi }{32}}}$

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$x^2 – 5x + 6 = (x – 2)(x – 3)$

Gradient = $\dfrac{11 – 3}{6 – 2} = \dfrac{8}{4} = 2$

$0.00843 = 8.43 \times 10^{-3}$

$x = \dfrac{3 \pm \sqrt{(-3)^2 + 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \dfrac{3 \pm \sqrt{49}}{2} = \dfrac{3 \pm 7}{2} \Rightarrow x = 5, -2$