Question

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 1

Answer:

Step-by-step explanation:

(a) $\int x^n e^{5x^4+1} \ dx$

Suppose $5x^4 + 1 = f$

by differentiation;

$\implies \ 20 x^3 dx = df --- (1)$

Suppose n = 3

Then, the integral

$I = \int x^ 3 e^{5x^4 + 1} \ dx$

$= \int e^f \ \dfrac{df}{20}$

$= \dfrac{1}{20} \int e^f \ dt$

$= \dfrac{1}{20} e^f + C$

recall that $f = 5x^4 + 1$

Then;

$\mathbf{ I = \dfrac{1}{20}e^{5x^4+1}+C}$

(b) $\int \dfrac{cos (\dfrac{1}{x^3})}{x^n } \ dx$

suppose; $\dfrac{1}{x^3} = f$

$x^3 = f$

$\implies -3x^{-4} \ dx = df$

$\implies \dfrac{1}{x^4} \ dx =-\dfrac{1}{3} df$

If n = u, then the integration is:

$I = \int \dfrac{1}{x^4} \ cos (\dfrac{1}{x^4}) \ dx$

$= \int -\dfrac{1}{3} \ cos \ f \ df$

$= -\dfrac{1}{3} \int \ cos \ f \ df$

$= -\dfrac{1}{3} \ sin \ f + C$

Since;  $x^3 = f$

Then;

$\mathbf {I = -\dfrac{1}{3} \ sin \ \Big( \dfrac{1}{x^3}\Big) + C}$

(c) $\int \dfrac{x+n}{x^2 + 8x -4} \ dx$

Suppose  $x^2 + 8x - 4 = f$

Then, by differentiation of both sides

$(2x + 8) \ dx = df$

$(x + 4) \ dx = \dfrac{1}{2} \ df$

Suppose n = 4 in integration, then:

$I = \int \dfrac{(x + 4) }{x^2 +8x -4} \ dx$

By substitution;

$I = \int \dfrac{1}{2}\dfrac{1}{f} \ df$

$= \dfrac{1}{2} \ \ { In |f|} + C$

$\mathbf{= \dfrac{1}{2} \ \ { In |x^2+8x -4|} + C}$