Answer:
[tex]\textsf{1)} \quad -\dfrac{1}{16}e^{-4x}\left(4x+1\right)+\text{C}[/tex]
[tex]\textsf{2)} \quad - \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}[/tex]
Step-by-step explanation:
Question 1
[tex]\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5 cm}\underline{Integration of $e^{ax}$} \\\\$\displaystyle \int e^{ax}\:\text{d}x=\dfrac{1}{a}e^{ax}+\text{C}$\\\\for $a\neq 0$\\\end{minipage}}[/tex]
Given integral:
[tex]\displaystyle \int xe^{-4x}\:\text{d}x[/tex]
Using Integration by parts:
[tex]\textsf{Let }\:u=x \implies \dfrac{\text{d}u}{\text{d}x}=1[/tex]
[tex]\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=e^{-4x} \implies v=-\dfrac{1}{4}e^{-4x}[/tex]
Therefore:
[tex]\begin{aligned}\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x & =uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x\\\\\implies \displaystyle \int xe^{-4x}\:\text{d}x & =-\dfrac{1}{4}xe^{-4x}-\int -\dfrac{1}{4}e^{-4x}\: \text{d}x\\\\& =-\dfrac{1}{4}xe^{-4x}+\int \dfrac{1}{4}e^{-4x}\: \text{d}x\\\\& =-\dfrac{1}{4}xe^{-4x}-\dfrac{1}{16}e^{-4x}+\text{C}\\\\& =-\dfrac{1}{16}e^{-4x}\left(4x+1\right)+\text{C}\end{aligned}[/tex]
Question 2
[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\\ \end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\(where $n$ is any constant value)\end{minipage}}[/tex]
Rewrite the given integral:
[tex]\begin{aligned}\displaystyle \int \sin^5 x \: \text{d}x & =\int (\sin x)^4 \cdot \sin x \: \text{d}x\\& =\int (\sin^2 x)^2 \cdot \sin x \: \text{d}x\end{aligned}[/tex]
Use the trig identity [tex]\sin^2x+\cos^2x \equiv 1[/tex] to rewrite [tex]\sin^2x[/tex] :
[tex]\implies \displaystyle \int \sin^5 x \: \text{d}x = \int (1-\cos^2 x)^2 \cdot \sin x \: \text{d}x[/tex]
Integration by substitution
[tex]\textsf{Let }\:u=\cos x \implies \dfrac{\text{d}u}{\text{d}x}=-\sin x \implies \text{d}x=-\dfrac{1}{\sin x}\: \text{d}u[/tex]
Therefore:
[tex]\begin{aligned}\implies \displaystyle \int \sin^5 x \: \text{d}x & = \int (1-u^2)^2 \cdot \sin x \cdot -\dfrac{1}{\sin x}\: \text{d}u\\& = \int -(1-u^2)^2 \: \text{d}u\\ & =\int -1+2u^2-u^4 \: \text{d}u\\& =-u+\dfrac{2}{3}u^3-\dfrac{1}{5}u^5+\text{C}\end{aligned}[/tex]
Finally, substitute [tex]u = \cos x[/tex] back in:
[tex]\implies \displaystyle \int \sin^5 x \: \text{d}x=- \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}[/tex]
