Exercise 10.2

Exercise 10.2: Integration by Substitution

1. Integrate the following functions using the given substitutions.

(a) \(\int 4x^3 \sqrt{x^4-1} \,dx; \quad u = x^4 - 1\)

Solution

\( u = x^4 - 1 \)

Then \( du = 4x^3 \,dx \)

\(\int 4x^3 \sqrt{x^4-1} \,dx = \int \sqrt{u} \,du = \int u^{\frac{1}{2}} \,du \)

\(= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C \)

\(= \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{3} (x^4 - 1)^{\frac{3}{2}} + C \)

(b) \(\int \cos^3 x \sin x \,dx; \quad u = \cos x\)

Solution

\( u = \cos x \)

Then \( du = -\sin x \,dx \implies -du = \sin x \,dx \)

\(\int \cos^3 x \sin x \,dx = \int u^3 (-du) \)

\(= -\int u^3 \,du \)

\(= -\frac{u^{3+1}}{3+1} + C \)

\(= -\frac{1}{4} u^4 + C \)

\(= -\frac{1}{4} \cos^4 x + C \)

(c) \(\int \frac{1}{x \ln|x|} \,dx; \quad u = \ln|x|\)

Solution

\( u = \ln|x| \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{1}{x \ln|x|} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\ln|x|| + C \)

(d) \(\int \sin^5 x \cos x \,dx; \quad u = \sin x\)

Solution

\( u = \sin x \)

Then \( du = \cos x \,dx \)

\(\int \sin^5 x \cos x \,dx = \int u^5 \,du \)

\(= \frac{u^{5+1}}{5+1} + C \)

\(= \frac{u^6}{6} + C \)

\(= \frac{\sin^6 x}{6} + C \)

(e) \(\int \frac{\ln x}{x} \,dx, x > 0; \quad u = \ln x\)

Solution

\( u = \ln x \)

Then \( du = \frac{1}{x} \,dx \)

\(\int \frac{\ln x}{x} \,dx = \int u \,du \)

\(= \frac{u^{1+1}}{1+1} + C \)

\(= \frac{1}{2} u^2 + C \)

\(= \frac{1}{2} (\ln x)^2 + C \)

(f) \(\int x^3 e^{x^4} \,dx; \quad u = x^4\)

Solution

\( u = x^4 \)

Then \( du = 4x^3 \,dx \implies \frac{1}{4} du = x^3 \,dx \)

\(\int x^3 e^{x^4} \,dx = \int e^u (\frac{1}{4} du) \)

\(= \frac{1}{4} \int e^u \,du \)

\(= \frac{1}{4} e^u + C \)

\(= \frac{1}{4} e^{x^4} + C \)

2. Use the substitution method to evaluate the following integrals.

(a) \(\int x \sqrt{1-x} \,dx\)

Solution

Let \( u = 1 - x \implies x = 1 - u \)

Then \( dx = -du \)

\(\int x \sqrt{1-x} \,dx = \int (1-u)\sqrt{u}(-du) \)

\(= \int (u-1)u^{\frac{1}{2}} \,du = \int (u^{\frac{3}{2}} - u^{\frac{1}{2}}) \,du \)

\(= \frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}} + C \)

\(= \frac{2}{5}(1-x)^{\frac{5}{2}} - \frac{2}{3}(1-x)^{\frac{3}{2}} + C \)

(b) \(\int (2x+1)(x^2+x)^7 \,dx\)

Solution

Let \( u = x^2+x \)

Then \( du = (2x+1) \,dx \)

\(\int (2x+1)(x^2+x)^7 \,dx = \int u^7 \,du \)

\(= \frac{1}{8} u^8 + C \)

\(= \frac{1}{8} (x^2+x)^8 + C \)

(c) \(\int \sin^3 x \,dx\)

Solution

\(\int \sin^3 x \,dx = \int \sin^2 x \cdot \sin x \,dx \)

\(= \int (1 - \cos^2 x) \sin x \,dx \)

Let \( u = \cos x \implies du = -\sin x \,dx \)

\(= \int (1-u^2) (-du) = \int (u^2-1) \,du \)

\(= \frac{1}{3}u^3 - u + C \)

\(= \frac{1}{3}\cos^3 x - \cos x + C \)

(d) \(\int x^2 \sqrt{x^3 - 2} \,dx\)

Solution

Let \( u = x^3 - 2 \)

Then \( du = 3x^2 \,dx \implies \frac{1}{3}du = x^2 \,dx \)

\(\int x^2 \sqrt{x^3 - 2} \,dx = \int \sqrt{u} (\frac{1}{3} du) \)

\(= \frac{1}{3} \int u^{\frac{1}{2}} \,du = \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} u^{\frac{3}{2}} + C \)

\(= \frac{2}{9} (x^3-2)^{\frac{3}{2}} + C \)

(e) \(\int \frac{\sec^2 x}{\tan x} \,dx\)

Solution

Let \( u = \tan x \)

Then \( du = \sec^2 x \,dx \)

\(\int \frac{\sec^2 x}{\tan x} \,dx = \int \frac{1}{u} \,du \)

\(= \ln|u| + C \)

\(= \ln|\tan x| + C \)

(f) \(\int \frac{x}{\sqrt{x+1}} \,dx\)

Solution

Let \( u = x+1 \implies x = u-1 \)

Then \( dx = du \)

\(\int \frac{x}{\sqrt{x+1}} \,dx = \int \frac{u-1}{\sqrt{u}} \,du \)

\(= \int (u^{\frac{1}{2}} - u^{-\frac{1}{2}}) \,du \)

\(= \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} + C \)

\(= \frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C \)

3. Evaluate the integral \(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx\).

Solution

Let \( u = \ln(x^2+1) \)

Then \( du = \frac{1}{x^2+1} \cdot 2x \,dx \implies \frac{1}{2}du = \frac{x}{x^2+1} \,dx \)

\(\int \frac{x}{(x^2+1)\ln(x^2+1)} \,dx = \int \frac{1}{u} (\frac{1}{2} du) \)

\(= \frac{1}{2} \int \frac{1}{u} \,du \)

\(= \frac{1}{2} \ln|u| + C \)

\(= \frac{1}{2} \ln|\ln(x^2+1)| + C \)