Calculate integrals using different techniques with examples and detailed solutions and explanations. More exercises with solutions are presented at the bottom of the page.
Note: In all examples and exercises below, \( c \) represents the constant of integration.
Examples and Their Detailed Solutions
Example 1: \( \int 6 \; \cos x \; \sin x \; dx \)
We first use the trigonometric identity \( 2 \sin x \cos x = \sin(2x) \) to rewrite the integral:
\[ \int 6 \; \cos x \; \sin x \; dx = 3 \int \sin(2x) \; dx \]Use Integration by Substitution: Let \( u = 2x \), which leads to \( \dfrac{du}{dx} = 2 \) or \( dx = du / 2 \):
\[ = 3 \int (1/2) \sin u \; du = - (3/2) \cos u + c \]Substitute back \( u = 2x \):
\[ \int 6 \cos x \sin x \, dx = - (3/2) \cos(2x) + c \]Check: differentiate \( - (3/2) \cos(2x) + c \) to obtain \( 6 \sin x \cos x \), the original integrand.
Example 2: \( \int x \; \sqrt{x+1} \, dx \)
Use Integration by Substitution: Let \( u = x + 1 \) which gives \( x = u - 1 \) and \( du = dx \).
\[ \int x \sqrt{x+1} \; dx = \int (u-1) \cdot u^{1/2} \; du = \int (u^{3/2} - u^{1/2}) \; du \]Use the power rule for integration (\( \int x^n dx = \dfrac{1}{n+1} x^{n+1} + c \)):
\[ = (2 / 5) u^{5/2} - (2 / 3) u^{3/2} + c \]Substitute back \( u = x + 1 \):
\[ \int x \sqrt{x+1} \, dx = (2 / 5) (x + 1)^{5/2} - (2 / 3) (x + 1)^{3/2} + c \]Example 3: \( \int \cos^2 x \; dx \)
Use the trigonometric identity \( \cos^2 x = \dfrac{1 + \cos(2x)}{2} \):
\[ \int \cos^2 x \; dx = \int \dfrac{1 + \cos(2x)}{2} \; dx \]Let \( u = 2x \) so that \( dx = du / 2 \):
\[ = \int (1/4)(1 + \cos u) \; du = (1 / 4) u + (1 / 4) \sin(u) + c \]Substitute back \( u = 2x \):
\[ \int \cos^2 x \; dx = x / 2 + (1 / 4) \sin(2x) + c \]Example 4: \( \int x^3 e^{x^4} \; dx \)
Let \( u = x^4 \), so \( (1 / 4) du = x^3 dx \):
\[ \int x^3 e^{x^4} dx = \int (1 / 4) e^u \; du = (1 / 4) e^u + c \]Substitute back \( u = x^4 \):
\[ \int x^3 e^{x^4} dx = (1 / 4) e^{x^4} + c \]Example 5: \( \int \dfrac{\sin(2x)}{1 - \cos^2(x)} \; dx \)
Use identities \( \sin(2x) = 2 \sin x \cos x \) and \( 1 - \cos^2(x) = \sin^2(x) \):
\[ \int \dfrac{2 \sin x \cos x}{\sin^2(x)} dx = \int 2 \dfrac{\cos x}{\sin x} \; dx \]Use the formula \( \int \dfrac{f'(x)}{f(x)} \; dx = \ln|f(x)| + c \):
\[ = 2 \ln|\sin x| + c \]Example 6: \( \int (x + \sin x)^2 \; dx \)
Expand \( (x + \sin x)^2 = x^2 + \sin^2 x + 2x \sin x \) and apply the sum rule:
\[ \int (x + \sin x)^2 \; dx = \int x^2 \; dx + \int \sin^2 x \; dx + \int 2x \sin x \; dx \]\( \int x^2 \; dx = (1/3) x^3 + c \)
Use \( \sin^2 x = (1 - \cos(2x)) / 2 \):
\( \int \sin^2 x \; dx = (1/2) x - (1/4) \sin(2x) + c \)
For \( \int 2x \sin x \; dx \), use integration by parts (\( v = x, u' = \sin x \)):
\[ \int 2x \sin x \, dx = -2x \cos x + 2 \sin x + c \]Summing them all up:
\[ \int (x + \sin x)^2 \; dx = (1/3) x^3 + (1/2) x - (1/4) \sin(2x) - 2x \cos x + 2 \sin x + c \]Example 7: \( \int \dfrac{\sin x}{\sin^2 x - 2 \cos x + 2} \; dx \)
Use \( \sin^2 x = 1 - \cos^2 x \). Let \( u = \cos x \), then \( du = - \sin x \; dx \):
\[ \int \dfrac{\sin x}{1 - \cos^2 x - 2 \cos x + 2} \; dx = \int \dfrac{1}{u^2 + 2u - 3} \, du \]Factor \( u^2 + 2u - 3 = (u+3)(u-1) \) and use partial fractions decomposition:
\[ \dfrac{1}{u^2 + 2u - 3} = \dfrac{1}{4(u-1)} - \dfrac{1}{4(u+3)} \]Integrate and substitute back \( u = \cos x \):
\[ = (1/4) (\ln|u-1| - \ln|u+3|) + c \] \[ \int \dfrac{\sin x}{\sin^2 x - 2 \cos x + 2} \; dx = (1/4) \left(\ln \left|\dfrac{\cos x - 1}{\cos x + 3}\right|\right) + c \]Example 8: \( \int \dfrac{1}{x + \sqrt{x+2}} \; dx \)
Let \( u = \sqrt{x + 2} \) giving \( x = u^2 - 2 \) and \( dx = 2u \; du \).
\[ \int \dfrac{1}{x + \sqrt{x+2}} \; dx = 2 \int \dfrac{u}{u^2 - 2 + u} \; du \]Factor \( u^2 + u - 2 = (u+2)(u-1) \) and use partial fractions:
\[ \dfrac{u}{u^2 + u - 2} = \dfrac{1}{3(u-1)} + \dfrac{2}{3(u+2)} \] \[ 2 \int \dfrac{u}{u^2 + u - 2} \; du = 2 \int \dfrac{1}{3(u-1)} du + 2 \int \dfrac{2}{3(u+2)} du \] \[ = (2/3) \ln|u-1| + (4/3) \ln|u+2| + c \]Substitute \( u = \sqrt{x + 2} \):
\[ \int \dfrac{1}{x + \sqrt{x+2}} \; dx = (2/3) \ln|\sqrt{x+2} - 1| + (4/3) \ln|\sqrt{x+2} + 2| + c \]Example 9: \( \int \dfrac{1}{\tan x} \; dx \)
Use the identity \( \tan(x) = \dfrac{\sin(x)}{\cos(x)} \):
\[ \int \dfrac{1}{\tan x} \; dx = \int \dfrac{\cos x}{\sin x} \; dx \]Use the formula \( \int \dfrac{f'(x)}{f(x)} \; dx = \ln|f(x)| + c \):
\[ \int \dfrac{1}{\tan x} \; dx = \ln|\sin x| + c \]Example 10: \( \int \dfrac{1}{x^2 + 2x + 1} \; dx \)
Complete the square in the denominator: \( x^2 + 2x + 1 = (x+1)^2 \)
\[ \int \dfrac{1}{(x+1)^2} \; dx \]Let \( u = x + 1 \) and \( dx = du \):
\[ = \int u^{-2} \; du = - \dfrac{1}{u} + c \]Substitute back \( u = x + 1 \):
\[ \int \dfrac{1}{x^2 + 2x + 1} \; dx = - \dfrac{1}{x+1} + c \]Example 11: \( \int \dfrac{1}{x^2 + x + 1} \; dx \)
Complete the square in the denominator: \( x^2 + x + 1 = (x + 1/2)^2 + 3/4 \)
Let \( u = x + 1/2 \), then factor out \( 3/4 \):
\[ \int \dfrac{1}{u^2 + 3/4} du = \dfrac{4}{3} \int \dfrac{1}{\dfrac{4}{3} u^2 + 1} du \]Substitute \( w = \dfrac{2}{\sqrt{3}} u \), yielding \( du = \dfrac{\sqrt{3}}{2} dw \):
\[ \dfrac{4}{3} \dfrac{\sqrt{3}}{2} \int \dfrac{1}{w^2 + 1} dw = \dfrac{2}{\sqrt{3}} \arctan w \]Substitute back \( w = \dfrac{2}{\sqrt{3}} (x + 1/2) \):
\[ \int \dfrac{1}{x^2 + x + 1} dx = \dfrac{2}{\sqrt{3}} \arctan\left(\dfrac{2}{\sqrt{3}} (x + 1/2)\right) + c \]Example 12: \( \int \dfrac{x^4 - 2x^2 + x}{x^2 + x + 1} \; dx \)
Divide the numerator by the denominator since the degree is greater:
\[ \dfrac{x^4 - 2x^2 + x}{x^2 + x + 1} = x^2 - x - 2 + \dfrac{4x + 2}{x^2 + x + 1} \]\( \dfrac{4x + 2}{x^2 + x + 1} \) can be written as \( 2 \dfrac{2x + 1}{x^2 + x + 1} \).
Because \( 2x + 1 \) is the derivative of \( x^2 + x + 1 \), we use the log rule:
\[ \int \dfrac{x^4 - 2x^2 + x}{x^2 + x + 1} dx = (1/3)x^3 - (1/2)x^2 - 2x + 2 \ln|x^2 + x + 1| + c \]Example 13: \( \int \left(x^3 - \dfrac{1}{x^2}\right)^4 \; dx \)
Use the binomial theorem to expand the integrand:
\[ \left(x^3 - \dfrac{1}{x^2}\right)^4 = x^{12} - 4x^7 + 6x^2 - \dfrac{4}{x^3} + \dfrac{1}{x^8} \]Use integration of a power function to get the final answer:
\[ \int \left(x^3 - \dfrac{1}{x^2}\right)^4 \; dx = \dfrac{x^{13}}{13} - \dfrac{x^8}{2} + 2x^3 + \dfrac{2}{x^2} - \dfrac{1}{7x^7} + c \]Example 14: \( \int \tan^2(x) \; dx \)
Use the identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integral:
\[ \int \tan^2(x) \; dx = \int (\sec^2 x - 1) \; dx = \int \sec^2 x \; dx - \int 1 \; dx \]Use the formula \( \int \sec^2 x dx = \tan(x) + c \):
\[ \int \tan^2(x) \; dx = \tan(x) - x + c \]Example 15: \( \int x^4(4x^5 - 2)^{10} \; dx \)
Let \( u = 4x^5 - 2 \) which gives \( du = 20x^4 dx \):
\[ \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{20} \int u^{10} du = \dfrac{1}{220} u^{11} + c \]Substitute back \( u = 4x^5 - 2 \):
\[ \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{220} ( 4x^5 - 2 )^{11} + c \]Example 16: \( \int x^2 \arcsin(x) \; dx \)
Use Integration by parts: Let \( w' = x^2 \) (\( w = (1/3) x^3 \)) and \( v = \arcsin(x) \) (\( v' = \dfrac{1}{\sqrt{1-x^2}} \)).
\[ \int x^2 \arcsin(x) \; dx = (1/3) x^3 \arcsin(x) - (1/3) \int x^3 \dfrac{1}{\sqrt{1-x^2}} \; dx \]Solve the right integral with substitution \( u = \sqrt{1-x^2} \) to obtain:
\[ \int x^2 \arcsin(x) \; dx = \dfrac{1}{3} x^3 \arcsin(x) - \dfrac{1}{3} \left( \dfrac{1}{3} (1-x^2)^{3/2} - \sqrt{1-x^2} \right) + c \]Example 17: \( \int \sqrt{x} \ln x \; dx \)
Use integration by parts. Let \( w' = \sqrt{x} \) (\( w = (2/3)x^{3/2} \)) and \( v = \ln x \) (\( v' = 1/x \)):
\[ \int \sqrt{x} \ln x \; dx = (2/3)x^{3/2} \ln x - \int (2/3)x^{1/2} \; dx \] \[ = \dfrac{2}{3} \; x^{3/2} \ln x - \left(\dfrac{2}{3}\right)^2 x^{3/2} + c \]Example 18: \( \int \dfrac{\sqrt{x+1}}{x} \; dx \)
Let \( u = \sqrt{x+1} \) making \( dx = 2u \; du \) and \( x = u^2 - 1 \):
\[ \int \dfrac{\sqrt{x+1}}{x} \; dx = \int \dfrac{u}{u^2 - 1} 2u \; du = 2 \int \dfrac{u^2}{u^2 - 1} \; du \]Divide the numerator and rewrite with partial fractions:
\[ \dfrac{u^2}{u^2 - 1} = 1 - \dfrac{1}{2(u+1)} + \dfrac{1}{2(u-1)} \] \[ 2 \int \left(1 - \dfrac{1}{2(u+1)} + \dfrac{1}{2(u-1)}\right) \; du = 2 \left(u - \dfrac{1}{2} \ln|u+1| + \dfrac{1}{2} \ln|u-1|\right) + c \]Substitute back \( u = \sqrt{x+1} \):
\[ \int \dfrac{\sqrt{x+1}}{x} \; dx = 2 \sqrt{x+1} + \ln \left(\dfrac{|x|}{(\sqrt{x+1} + 1)^2}\right) + c \]Example 19: \( \int \sin\left(\sqrt{x}\right) \; dx \)
Let \( u = \sqrt{x} \), \( dx = 2u \; du \):
\[ \int \sin\left(\sqrt{x}\right) \; dx = 2 \int u \sin u \; du \]Apply integration by parts to \( \int u \sin u \; du \):
\[ \int u \sin u \; du = -u \cos u + \int \cos u \, du = -u \cos u + \sin u + c \]Back substitute \( u = \sqrt{x} \):
\[ \int \sin\left(\sqrt{x}\right) \; dx = -2 \sqrt{x} \cos\sqrt{x} + 2 \sin\sqrt{x} + c \]Example 20: \( \int \dfrac{1}{e^x + e^{-x}} \; dx \)
Let \( u = e^x \), \( dx = \dfrac{1}{u} du \):
\[ \int \dfrac{1}{e^x + e^{-x}} \; dx = \int \dfrac{1}{u + 1/u} \; \dfrac{1}{u} \; du = \int \dfrac{1}{u^2 + 1} \; du \]Use the common integral \( \int \dfrac{1}{u^2+1}du = \arctan(u) \):
\[ \int \dfrac{1}{e^x + e^{-x}} \; dx = \arctan(e^x) + c \]Example 21: \( \int \log_5 x \; dx \)
Use the change of base formula \( \log_5(x) = \dfrac{\ln x}{\ln 5} \):
\[ \int \log_5 x \; dx = \dfrac{1}{\ln 5} \int \ln x \; dx \]Apply integration by parts to \( \int \ln x \, dx \) (\( w' = 1, v = \ln x \)):
\[ \int \ln x \; dx = x \ln x - \int x (1/x) \; dx = x \ln x - x + c \] \[ \int \log_5 x \; dx = \dfrac{1}{\ln 5} (x \ln x - x) + c \]Example 22: \( \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx \)
Use trigonometric substitution: Let \( \dfrac{x}{4} = \sin t \) which gives \( x = 4 \sin t \) and \( dx = 4 \cos t \, dt \):
\[ \sqrt{16 - x^2} = 4 \sqrt{1 - \sin^2 t} = 4 \cos t \] \[ \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = \int \dfrac{(4 \sin t)^2}{4 \cos t} \; 4 \cos t \, dt = 16 \int \sin^2 t \; dt \]Use \( \sin^2 t = (1/2)(1 - \cos(2t)) \):
\[ 16 \int (1/2)(1 - \cos(2t)) \; dt = 8t - 4 \sin(2t) + c \]Substitute \( t = \arcsin(x/4) \) and simplify:
\[ \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 8 \arcsin(x/4) - \dfrac{x\sqrt{16 - x^2}}{2} + c \]Practice Exercises
Use the table of integrals and the techniques demonstrated to calculate the following integrals. Expand the boxes to check your solutions.