TB. 2 - Calculus

Question

Review of integration by substitution: Use u-substitution to find each of the following integrals.

a. $\int e^{3x+1} dx$ b. $\int x{e}^{x^2+1} dx$

c. $\int \frac{\ln x}{x} dx$ d. $\int \frac{1}{x \ln x} dx$

e. $\int \tan x {\sec }^{2}x dx$ f. $\int \frac{1}{x^2+1} dx$

g.$\int \sin x \sqrt{\cos x} dx$ h. $\int e^x \sin e^x dx$

Step-by-Step Solution

Answer

Part (a) $\int e^{3x+1} dx = \frac{e^{3x+1}}{3} + C$

Part (b) $\int x{e}^{x^2+1} dx = \frac{e^{x^2+1}}{2} + C$

Part (c) $\int \frac{\ln x}{x} dx = \frac{{\left(\ln x\right)}^2}{2} + C$

Part (d) $\int \frac{1}{x \ln x} dx = \ln \left|\ln x\right| + C$

Part (e) $\int \tan x {\sec }^{2}x dx = \frac{{\tan }^{2 }x}{2} + C$

Part (f) $\int \frac{1}{x^2+1} dx = {\tan }^{-1}x + C$

Part (g) $\int \sin x \sqrt{\cos x} dx = -\frac{2 {\cos }^{{\displaystyle \frac{3}{2}}}x}{3} + C$

Part (h) $\int e^x \sin e^x dx = -\cos \left(e^x\right) + C$

Part (a) Step 1. Calculating the integral

Given integral, $\int e^{3x+1} dx$

Substituting, $3x+1=u, 3dx=du$

$$\begin{gathered} \int \frac{e^u }{3}du =\frac{ e^u}{3}+ C \\ =\frac{e^{3x+1}}{3} + C \end{gathered}$$

Part (b) Step 1. Calculating the integral

Given integral, $\int x{e}^{x^2+1} dx$

Substituting, $x^2+1=u, 2x dx=du$

$$\begin{gathered} \int \frac{e^u}{2} du = \frac{e^u}{2} + C \\ =\frac{e^{x^2+1}}{2} + C \end{gathered}$$

Part (c) Step 1. Calculating the integral

Given integral, $\int \frac{\ln x}{x} dx$

Substituting, $\ln x=u, \frac{dx}{x}=du$

$$\begin{gathered} \int u du =\frac{u^2}{2} + C \\ =\frac{{\left(\ln x\right)}^2}{2} + C \end{gathered}$$

Part (d) Step 1. Calculating the integral

Given integral, $\int \frac{1}{x \ln x} dx$

Substituting, $\ln x = u, \frac{dx}{x}=du$

$$\begin{gathered} \int \frac{1}{u} du =\ln \left|u\right| + C \\ =\ln \left|\ln x\right| + C \end{gathered}$$

Part (e) Step 1. Calculating the integral

Given integral, $\int \tan x {\sec }^{2}x dx$

Substituting, $\tan x=u, se{c}^{2}x dx=du$

$$\begin{gathered} \int u du = \frac{u^2}{2} + C \\ =\frac{{\tan }^{2}x}{2} + C \end{gathered}$$

Part (f) Step 1. Calculating the integral

Given integral, $\int \frac{1}{x^2+1} dx$

$={\tan }^{-1}x + C$

Part (g) Step 1. Calculating the integral

Given integral, $\int \sin x \sqrt{\cos x} dx$

Substituting, $\cos x = u, -\sin x dx = du$

$$\begin{gathered} -\int \sqrt{u} du =-\frac{u^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}} + C \\ =-\frac{2 {\cos }^{{\displaystyle \frac{3}{2}}}x}{3} + C \end{gathered}$$

Part (h) Step 1. Calculating the integral

Given integral, $\int e^x \sin e^x dx$

Substituting, $e^x=u, e^{x}dx=du$

$$\begin{gathered} \int \sin u du =-\cos u + C \\ =-\cos \left(e^x\right) + C \end{gathered}$$