Integration techniques: VJC 2024 Promo Q10

(a)(i)$\displaystyle\frac{d}{dx}(e^{\tan x})=e^{\tan x}\sec^2x$

10(a)(ii)$\displaystyle\int e^{\tan x}\sec^3x\sin x\,dx=\int(e^{\tan x}\sec^2x)\tan x\,dx$

$$=e^{\tan x}\tan x-\int e^{\tan x}\sec^2x\,dx=e^{\tan x}\tan x-e^{\tan x}+C$$ Side working:

$u=\tan x\Rightarrow\dfrac{du}{dx}=\sec^2x$

$\dfrac{dv}{dx}=e^{\tan x}\sec^2x\Rightarrow v=e^{\tan x}$

10(b)$1-4x=A(3-8x)+B$

By observation, $-8A=-4\Rightarrow A=\dfrac12$

$3A+B=1\Rightarrow B=1-\dfrac32=-\dfrac12$

$$\int\frac{1-4x}{\sqrt{1-4x^2+3x}}dx=\frac12\int\frac{3-8x}{\sqrt{1-4x^2+3x}}dx-\frac12\int\frac1{\sqrt{1-4x^2+3x}}dx$$ $$=\frac12(2)\sqrt{1-4x^2+3x}-\frac12\sin^{-1}\left(\frac{x-\frac38}{\frac58}\right)+C$$ $$=\sqrt{1-4x^2+3x}-\frac14\sin^{-1}\left(\frac{8x-3}{5}\right)+C$$ 10(c)$u=\sqrt{1+x}\Rightarrow x=u^2-1$, $\dfrac{du}{dx}=\dfrac1{2\sqrt{1+x}}=\dfrac1{2u}$

When $x=3$, $u=2$. When $x=1$, $u=\sqrt2$.

$$\int_1^3\frac{x^2}{(1+x)^{3/2}}dx=\int_{\sqrt2}^{2}\frac{(u^2-1)^2(2u)}{u^3}du=2\int_{\sqrt2}^{2}\left(u^2-2+\frac1{u^2}\right)du$$ $$=2\left[\frac{u^3}{3}-2u-\frac1u\right]_{\sqrt2}^{2}=2\left[\left(\frac83-4-\frac12\right)-\left(\frac{2\sqrt2}{3}-2\sqrt2-\frac{\sqrt2}{2}\right)\right]=\frac{11}{3}(\sqrt2-1)$$