U-substitutions only work for specific kinds of expressions. Below, you are asked to choose a value of n for which u-substitutions will be a useful integration technique. Then, you are to compute the antiderivative with that specific n. (E.g., if n = 5 makes u-subs work, then solve the integral with a 5 in place of n).
(a) [ zºeke*+1 "'de
(b) /co cos(1/2) dr
(c) / r+n dr 22 + 8x - 4

Answer:

Step-by-step explanation:

(a) [tex]\int x^n e^{5x^4+1} \ dx[/tex]

Suppose [tex]5x^4 + 1 = f[/tex]

by differentiation;

[tex]\implies \ 20 x^3 dx = df --- (1)[/tex]

Suppose n = 3

Then, the integral

[tex]I = \int x^ 3 e^{5x^4 + 1} \ dx[/tex]

[tex]= \int e^f \ \dfrac{df}{20}[/tex]

[tex]= \dfrac{1}{20} \int e^f \ dt[/tex]

[tex]= \dfrac{1}{20} e^f + C[/tex]

recall that [tex]f = 5x^4 + 1[/tex]

Then;

[tex]\mathbf{ I = \dfrac{1}{20}e^{5x^4+1}+C}[/tex]

(b) [tex]\int \dfrac{cos (\dfrac{1}{x^3})}{x^n } \ dx[/tex]

suppose; [tex]\dfrac{1}{x^3} = f[/tex]

[tex]x^3 = f[/tex]

[tex]\implies -3x^{-4} \ dx = df[/tex]

[tex]\implies \dfrac{1}{x^4} \ dx =-\dfrac{1}{3} df[/tex]

If n = u, then the integration is:

[tex]I = \int \dfrac{1}{x^4} \ cos (\dfrac{1}{x^4}) \ dx[/tex]

[tex]= \int -\dfrac{1}{3} \ cos \ f \ df[/tex]

[tex]= -\dfrac{1}{3} \int \ cos \ f \ df[/tex]

[tex]= -\dfrac{1}{3} \ sin \ f + C[/tex]

Since;  [tex]x^3 = f[/tex]

Then;

[tex]\mathbf {I = -\dfrac{1}{3} \ sin \ \Big( \dfrac{1}{x^3}\Big) + C}[/tex]

(c) [tex]\int \dfrac{x+n}{x^2 + 8x -4} \ dx[/tex]

Suppose  [tex]x^2 + 8x - 4 = f[/tex]

Then, by differentiation of both sides

[tex](2x + 8) \ dx = df[/tex]

[tex](x + 4) \ dx = \dfrac{1}{2} \ df[/tex]

Suppose n = 4 in integration, then:

[tex]I = \int \dfrac{(x + 4) }{x^2 +8x -4} \ dx[/tex]

By substitution;

[tex]I = \int \dfrac{1}{2}\dfrac{1}{f} \ df[/tex]

[tex]= \dfrac{1}{2} \ \ { In |f|} + C[/tex]

[tex]\mathbf{= \dfrac{1}{2} \ \ { In |x^2+8x -4|} + C}[/tex]

The suitable substitutions of n are 3,4,4 respectively.

The process of finding integrals is called integration.

a)[tex]f(x)=\int\limits {x^3e^{5x^4+1} } \, dx[/tex]

Suppose

[tex]5x^4+1 =t\\20x^3 dx =dt[/tex]

So, we need n=3 for easy integration.

[tex]f(x)=\int\limits {x^3e^{5x^4+1} } \, dx[/tex]

[tex]I = \frac{1}{20} \int\limits {e^{t} } \, dt[/tex]

[tex]I=\frac{e^{t} }{20}[/tex]

[tex]I = e^{5x^{4}+1 }/20 +c[/tex]

b)Similarly for [tex]f(x) = \int\limits\frac{cos(\frac{1}{x^3} )}{x^n} \, dx[/tex]

n=4 is needed for easy integration.

I = [tex]\frac{-1}{3} sin(\frac{1}{x^3} ) +c[/tex]

c)For [tex]f(x) = \int\limits \frac{x+n}{x^{2} +8x-4} \, dx[/tex]

n=4 is needed for easy integration.

[tex]I = \frac{1}{2} log(x^{2} +8x-4)[/tex]

Hence, the suitable substitutions of n are 3,4,4 respectively.

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