<p>State the derivative of $\tan {{x}^{2}}$. Hence, or otherwise, find $\int{{{x}^{3}}{{\sec }^{2}}{{x}^{2}}\mathrm{d}x}$</p><p style="text-align: right;">[4]</p>

State the derivative of $\tan {{x}^{2}}$. Hence, or otherwise, find $\displaystyle\displaystyle \int{{{x}^{3}}{{\sec }^{2}}{{x}^{2}}\mathrm{d}x}$

Techniques of  Integration - General

Techniques of  Integration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

2020 EJC P1 Q2

$\frac{\mathrm{d}}{\mathrm{d}x}(\tan x^2) = 2x \sec^2(x^2)$

Using integration by parts: $\displaystyle\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \displaystyle\int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x$\nLet $u = x^2$, $\frac{\mathrm{d}v}{\mathrm{d}x} = x \sec^2 x^2$\n$\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$, $v = \frac{1}{2}\tan x^2$

$\displaystyle\int x^3 \sec^2 x^2 \, \mathrm{d}x = \displaystyle\int x^2 (x \sec^2 x^2) \, \mathrm{d}x$\n$= x^2 \cdot \frac{1}{2}\tan x^2 - \displaystyle\int \frac{1}{2}\tan x^2 \cdot 2x \, \mathrm{d}x$\n$= \frac{1}{2}x^2 \tan x^2 - \displaystyle\int x \tan x^2 \, \mathrm{d}x$

Let $k = x^2$, $\frac{dk}{\mathrm{d}x} = 2x$\n$= \frac{1}{2}x^2 \tan x^2 - \displaystyle\int x \tan k \cdot \frac{dk}{2x}$\n$= \frac{1}{2}x^2 \tan x^2 - \frac{1}{2}\displaystyle\int \tan k \, dk$

$= \frac{1}{2}x^2 \tan x^2 - \frac{1}{2}\ln(\sec k) + C$\n$= \boxed{\frac{1}{2}x^2 \tan x^2 - \frac{1}{2}\ln(\sec x^2) + C}$

2020 EJC P1 Q2

2020 EJC P1 Q2

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