Detailed Examples, Techniques, and Solutions for Calculus
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.
We first use the trigonometric identity \( 2 \sin x \cos x = \sin (2 x) \) to rewrite the integral:
\[ \int 6 \; \cos x \; \sin x \; dx = 3 \int \sin (2x) dx \]Use Integration by Substitution: Let \( u = 2 x \), 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 \( u = 2x \):
\[ \int 6 \cos x \sin x dx = - (3/2) \cos (2 x) + c \]Check: differentiate \( - (3/2) \cos (2x) + c \) to obtain \( 6 \sin x \cos x \), the original integrand.
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 \( u = x + 1 \):
\[ \int x \sqrt{x+1} dx = (2 / 5) (x + 1)^{5/2} - (2 / 3) (x + 1)^{3/2} + c \]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 \( u = 2 x \):
\[ \int \cos^2 x \; dx = x / 2 + (1 / 4) \sin (2 x) + c \]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 \( u = x^4 \):
\[ \int x^3 e^{x^4} dx = (1 / 4) e^{x^4}+ c \]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 \( \int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \):
\[ = 2 \ln |\sin x| + c \]Expand \( (x+\sin x)^2 = x^2 + \sin^2 x + 2 x \sin x \) and apply the sum rule:
\[ \int (x+\sin x)^2 \; dx = \int x^2 \; dx + \int \sin^2 x \; dx + \int 2 x \; \sin x \; dx \]\( \int x^2 \; dx = (1/3) x^3 + c \)
Use \( \sin^2 x = (1 – \cos (2x)) / 2 \):
\( \int \sin^2 x \; dx = (1/2) x - (1/4) \sin (2x) + c \)
For \( \int 2 x \sin x \; dx \), use integration by parts (\( v = x, u' = \sin x \)):
\[ \int 2 x \sin x dx = - 2 x \cos x + 2 \sin x + c \]Summing them all up:
\[ \int (x+\sin x)^2 \; dx = (1/3) x^3 + (1/2) x - (1/4) \sin (2x) - 2 x \cos x + 2 \sin x + c \]Use \( \sin^2 x = 1 - \cos^2 x \). Let \( u = \cos x \), then \( du = - \sin x \; dx \):
\[ \int \dfrac{\sin x}{1 - \cos^2 x - 2 \cos x + 2} \; dx = \int \dfrac{1}{u^2 + 2 u - 3} du \]Factor \( u^2 + 2 u - 3 = (u+3)(u-1) \) and use partial fractions decomposition:
\[ \dfrac{1}{u^2 + 2 u - 3} = \dfrac{1}{4(u-1)} - \dfrac{1}{4(u+3)} \]Integrate and substitute back \( u = \cos x \):
\[ = (1/4) (\ln |u-1| - \ln|u+3|) + c \] \[ \int \dfrac{\sin x}{\sin^2 x - 2 \cos x + 2} \; dx = (1/4) \left(\ln \left|\dfrac{ \cos x -1}{\cos x +3} \right|\right) + c \]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 \( u = \sqrt {x + 2} \):
\[ \int \dfrac{1}{x+\sqrt{x+2}} \; dx = (2/3) \ln |\sqrt {x + 2}-1| + (4/3) \ln|\sqrt {x + 2}+2| + c \]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 \( \int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \):
\[ \int \dfrac{1}{\tan x} \; dx = \ln | \sin x | + c \]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 \( u = x+ 1 \):
\[ \int \dfrac{1}{x^2 + 2x + 1} \; dx = - \dfrac{1}{x+1} + c \]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 \( w = \dfrac{2}{\sqrt 3} (x+1/2) \):
\[ \int \dfrac{1}{x^2+x+1}dx = \dfrac{2}{\sqrt{3}} \arctan \left(\dfrac{2}{\sqrt 3} (x+1/2) \right) + c \]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:
\[ \int \dfrac{x^4-2x^2+x}{x^2+x+1}dx = (1/3)x^3 - (1/2)x^2 - 2 x + 2 \ln |x^2+x+1| + c \]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:
\[ \int \left(x^3-\dfrac{1}{x^2}\right)^4 \; dx = \dfrac{x^{13}}{13}-\dfrac{x^8}{2}+2x^3+\dfrac{2}{x^2}-\dfrac{1}{7x^7}+ c \]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 \( \int \sec^2 x dx = \tan (x) + c \):
\[ \int \tan^2(x) \; dx = \tan (x) - x + c \]Let \( u = 4x^5 - 2 \) which gives \( du = 20 x^4 dx \):
\[ \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{20} \int u^{10} du = \dfrac{1}{120} u^{11} + c \]Substitute back \( u = 4x^5 - 2 \):
\[ \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{240} ( 4x^5 - 2 )^{11} + c \]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:
\[ \int x^2 \arcsin(x) \; dx = \dfrac{1}{3} x^3 \arcsin(x) - \dfrac{1}{3} \left( \dfrac{1}{3} (1-x^2)^{3/2} - \sqrt {1-x^2} \right) + c \]Use integration by parts. Let \( w' = \sqrt x \) (\( w = (2/3)x^{3/2} \)) and \( v = \ln x \) (\( v' = 1/x \)):
\[ \int \sqrt x \ln x \; dx = (2/3)x^{3/2} \ln x - \int (2/3)x^{1/2} \; dx \] \[ = \dfrac{2}{3} \; x^{3/2} \ln x - \left(\dfrac{2}{3}\right)^2 x^{3/2} + c \]Let \( u = \sqrt{x+1} \) making \( dx = 2u \; du \) and \( x = u^2 - 1 \):
\[ \int \dfrac{\sqrt{x+1}}{x} \; dx = \int \dfrac{u}{u^2 - 1} 2 u \; du = 2 \int \dfrac{u^2}{u^2 - 1} \; du \]Divide the numerator and rewrite with partial fractions:
\[ \dfrac{u^2}{u^2 - 1} = 1 - \dfrac{1}{2\left(u+1\right)}+\dfrac{1}{2\left(u-1\right)} \] \[ 2 \int (1 - \dfrac{1}{2\left(u+1\right)}+\dfrac{1}{2\left(u-1\right)}) \; du = 2 (u - \dfrac{1}{2} \ln |u + 1 | + \dfrac{1}{2} \ln |u - 1| ) + c \]Substitute back \( u = \sqrt{x+1} \):
\[ \int \dfrac{\sqrt{x+1}}{x} \; dx = 2 \sqrt{x+1} + \ln \left(\dfrac{|x|}{(\sqrt{x+1} + 1)^2}\right) + c\]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 \( u = \sqrt{x} \):
\[ \int \sin\left(\sqrt{x}\right) \; dx = - 2 \sqrt{x} \cos \sqrt{x} + 2 \sin \sqrt{x} + c \]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 \( \int \dfrac{1}{u^2+1}du=\arctan \left(u\right) \):
\[ \int \dfrac{1}{e^x+e^{-x}} \; dx = \arctan e^x + c \]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 \( \int \ln x dx \) (\( w' = 1, v = \ln x \)):
\[ \int \ln x \; dx = x \ln x - \int x (1/x) \; dx = x \ln x - x + c \] \[ \int \log_5 x \; dx = \dfrac{1}{\ln 5} ( x \ln x - x) + c \]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 \( \sin^2 t = (1/2)(1 - \cos (2t)) \):
\[ 16 \int (1/2)(1- \cos (2t)) \; dt = 8 t - 4 \sin (2t) + c \]Substitute \( t = \arcsin(x/4) \) and simplify:
\[ \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 8 \arcsin(x/4) - \dfrac{x\sqrt{16-x^2}}{2} + c\]Use the table of integrals and the techniques demonstrated to calculate the following integrals. Expand the boxes to check your solutions.