Question

Please help me answer this question

Answer 1

Explanation:

look at the attachment above

Answer 2

`\textsf{1)} \quad -(1)/(16)e^(-4x)\left(4x+1\right)+\text{C}`

`\textsf{2)} \quad - \cos x+(2)/(3) \cos^3 x - (1)/(5) \cos^5 x +\text{C}`

Explanation:

Question 1

`\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \frac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \frac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}`

`\boxed{\begin{minipage}{5 cm}\underline{Integration of $e^(ax)$} \\\\$\displaystyle \int e^(ax)\:\text{d}x=(1)/(a)e^(ax)+\text{C}$\\\\for $a\\eq 0$\\\end{minipage}}`

Given integral:

`\displaystyle \int xe^(-4x)\:\text{d}x`

Using Integration by parts:

`\textsf{Let }\:u=x \implies \frac{\text{d}u}{\text{d}x}=1`

`\textsf{Let }\:\frac{\text{d}v}{\text{d}x}=e^(-4x) \implies v=-(1)/(4)e^(-4x)`

Therefore:

`\begin{aligned}\displaystyle \int u \frac{\text{d}v}{\text{d}x}\:\text{d}x & =uv-\int v\: \frac{\text{d}u}{\text{d}x}\:\text{d}x\\\\\implies \displaystyle \int xe^(-4x)\:\text{d}x & =-(1)/(4)xe^(-4x)-\int -(1)/(4)e^(-4x)\: \text{d}x\\\\& =-(1)/(4)xe^(-4x)+\int (1)/(4)e^(-4x)\: \text{d}x\\\\& =-(1)/(4)xe^(-4x)-(1)/(16)e^(-4x)+\text{C}\\\\& =-(1)/(16)e^(-4x)\left(4x+1\right)+\text{C}\end{aligned}`

Question 2

`\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=(x^(n+1))/(n+1)+\text{C}$\\ \end{minipage}}`

`\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}}`

Rewrite the given integral:

`\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}`

Use the trig identity
`\sin^2x+\cos^2x \equiv 1` to rewrite
`\sin^2x` :

`\implies \displaystyle \int \sin^5 x \: \text{d}x = \int (1-\cos^2 x)^2 \cdot \sin x \: \text{d}x`

Integration by substitution

`\textsf{Let }\:u=\cos x \implies \frac{\text{d}u}{\text{d}x}=-\sin x \implies \text{d}x=-(1)/(\sin x)\: \text{d}u`

Therefore:

`\begin{aligned}\implies \displaystyle \int \sin^5 x \: \text{d}x & = \int (1-u^2)^2 \cdot \sin x \cdot -(1)/(\sin x)\: \text{d}u\\& = \int -(1-u^2)^2 \: \text{d}u\\ & =\int -1+2u^2-u^4 \: \text{d}u\\& =-u+(2)/(3)u^3-(1)/(5)u^5+\text{C}\end{aligned}`

Finally, substitute
`u = \cos x` back in:

`\implies \displaystyle \int \sin^5 x \: \text{d}x=- \cos x+(2)/(3) \cos^3 x - (1)/(5) \cos^5 x +\text{C}`