Question

Please help me answer this question

Answer 1

Answer:

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

$\textsf{2)} \quad - \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}$

Step-by-step explanation:

Question 1

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

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

Given integral:

  $\displaystyle \int xe^{-4x}\:\text{d}x$

Using Integration by parts:

$\textsf{Let }\:u=x \implies \dfrac{\text{d}u}{\text{d}x}=1$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=e^{-4x} \implies v=-\dfrac{1}{4}e^{-4x}$

Therefore:

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

Question 2

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

$\boxed{\begin{minipage}{5 cm}\underline{Integrating a constant}\\\\ \displaystyle \int n\:\text{d}x=nx+\text{C} \\(where n is any constant value)\end{minipage}}$

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

$\textsf{Let }\:u=\cos x \implies \dfrac{\text{d}u}{\text{d}x}=-\sin x \implies \text{d}x=-\dfrac{1}{\sin x}\: \text{d}u$

Therefore:

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

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

$\implies \displaystyle \int \sin^5 x \: \text{d}x=- \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}$

Answer 2

Step-by-step explanation:

look at the attachment above