Indefinite Integration Questions and Answers

Use a substitution to evaluate the definite integral cos x da Osin 1 Ocos 1 sin 1

Given S is a surface for a part of the sphere x y z 4 that lies above the cone z x y Find a parametric representation for the surface S in term of o and 8 Then determine the domain of and 0 Based on the expression written previously show that rx rell 4 sin o Hence compute the surface integral of ffs ds where the surface S is given parametrically by r 0 0 Let C is the boundary curve along the bottom of the upper half of the sphere thus use Stokes theorem to compute the line integral of fo F dr where F x y z xzi yzj xyk and upwards orientation

Differentiate f x sin x cot x 4 F x cos x csc 2 x

Use integration by parts together with the techniques of this section to evaluate the integral Use C for the constant of integration x tan tan x dx

3 Use Log Rule to integrate a dx 6x 5 b 1 das

5 Find the area of the region that lies outside the circle r 3 sin 0 and inside the limacon r 1 2 sin

In a Use a substitution to find f Show your work and final answer in the provided area using HTML Editor s Insert Math Equation function an da

Use a substitution to find f sin x cos xdx O cos zsin x C 6 sin z cos x C OCCO E C in 2 C

Applying these values we get x dx 18 6 1 1 In x9 x18

The Integral Part 2 of 5 To find 1 X18 x dx Vx18 6 TV 6 can be best matched by formula number 43 we can use formula 43 shown below du u a2 In u u a C Using this we have u and a 43 from the Table of Integr

Part 5 of 5 x sin 4x2 cos 9x2 dx 20 Since cos 0 cos 8 then we can rewrite this as C 20 cos 13x 52 52 C

Applying these values we get 1 x sin 4x cos 9x dx cos 5x cos 13x C

Use the Table of Integrals to evaluate the integral sin 20 V 5 3 sin 8 C de

Use the Table of Integrals to evaluate the integral 1 x dx x18 6 Part 1 of 5 The integral J dx x18 6 can be best matched by formula number from the Table of Integrals Hint Note that x18 9

Use the Table of Integrals to evaluate the integral 9 si 9 sin7 x cos x In sin x dx

Use the Table of Integrals to evaluate the integral 21 sin 21 sin x dx Enhanced Feedback C Please try again Try using substitution along with the integral Ju u sin u du 2u 1 sin 4

Use the Table of Integrals to evaluate the integral 1 21 sin x dx X C

298 sin 80 de we can use formula 98 shown below Je Using this we have a To find eau sin bu du eau a b a sin bu b cos bu C I b u 0 and du d0

To find tan 16 x dx we can use formula 69 shown below tan u du tan u Inicos ul C To use this we ll let u Submit Skip you cannot come back Submit Answer and du dx

Evaluate the integral 1 t sin 4t dt C

Evaluate w xz dV where W is the domain bounded by the elliptic cylinder the first octant W Use symbolic notation and fractions where needed C xz dV II and the sphere x y z

Calculate the following antiderivatives a 13 b 13t5t 2 dt S c 1 u9 4 1 504 4 u du C dx C C

Find the average value of f x 00 00 X x 1 X 1 x 3 4 As

Step 4 With the substitution u sin x we get 15 sin x cos x dx 15 S which integrates to 13 5u 573 u du G Substituting back in to get the answer in terms of sin x we have sin x In terms of u we get 3u 15 sin x cos x dx 5 sin Step 5 Using the same substitution u sin x enables us to do 15 sin x cos x dx 15 Step 6 Combining these results we have J 15 sin x cos x dx 5 u du 3u 5 sin a C sin x C which in terms of sin x becomes 3 sin x 3 sin x C sin r and du cos x cos x dx

gives us Step 3 15 sin x 1 sin x cos x dx 15 sin x cos x dx Since cos x is the derivative of sin x then S 15 sin x cos x dx can be done by substituting u sin x Step 4 With the substitution u sin x we get 15 0 15 which integrates to 15 sin x cos x dx 15 5u u du C 5u Substituting back in to get the answer in terms of sin x we have 15 sin x cos x dx 5 sin x 5 sin x C In terms of u we get Step 5 Using the same substitution u sin x enables us to do f 15 15 sin x cos x dx 15 5 04 S 15 sin x cos x u du C which in terms of sin x becomes C 15 sin r cos r dx sin x and du cos x cos x dx

Since We know that 15 sin x cos x dx has an odd power of of cos x we will convert all but one power to sines cos x Step 2 Making this substitution using 15 sin x cos x dx gives us Step 3 15 sin x 1 sin x cos x dx 15 sin x cos x dx sin x Since cos x is the derivative of sin x then Step 4 With the substitution u sin x we get 15 0 which integrates to 15 sin x cos x dx 15 C 15 15 sin x cos x dx can be done by substituting u sin x u du Substituting back in to get the answer in terms of sin x we have 15 sin x cos x dx 15 sin x cos x C 15 sin r cos r dx sin r and du cos x cos x dx

Consider the following function 3 JO R 64 dx a Determine an appropriate trigonometric substitution Use x 8 sin 0 where 1 0 1 since the integrand contains the expression 82 x 2 Use x 8 tan 8 where x3 64 x IT 2 0x dx S 0 O Use x 8 sec 0 where 0 0 0 or 0 31 TT 2 TT since the integrand contains the expression 8 x 2 b Apply the substitution to transform the integral into a trigonometric integral Do not evaluate the integral 8 sin 0 8 cos 0 X since the integrand contains the expression 8 x de

With the substitution u sin x we get J which integrates to 12 sin x cos x dx 12 u du Substituting back in to get the answer in terms of sin x we have I 12 sin x cos x dx 4 sin x 4 sin r C 12 sin 4u Step 5 Using the same substitution u sin x enables us to do 12 u dur 12 sin x cos x dx 12 In terms of u we get 12 C Step 6 Combining these results we have 12 sin x cos x dx 125 5 C which in terms of sin x becomes 12 sin r 5 C

Since u 2 3x then converting back to an expression in x we get the following Remember to use absolute values where appropriate Inl Injul C

Since u 8 x then converting back to an expression in x we get 1 c 28 C

Now if u 8 x4 then This evaluates as x s dx us du 1 us du 8 x4 6 dx

x 18 We must also convert x dx into an expression involving u If u 8 x is substituted into x 8 x4 6 dx then we have We know that du 4x dx and so x dx du x u dx x x3 U6 x dx

Evaluate the integr l by making the given substitution Je 4x 4x dx U 4x

Consider the following function x x 3 a Determine an appropriate trigonometric substitution O Use x 3 sin 0 where 1505 12 24 0 since the integrand contains the expression x 3 I 2 dx O Use x 3 tan 0 where 3 T dx T O Use x 3 sec 0 where 0 0 or 0 2 TC 2 0 since the integrand contains the expression x2 3 V 2 b Apply the substitution to transform the integral into a trigonometric integral Do not evaluate the integral x x2 de 37 since the integrand contains the expression x2 3 2

Consider the following function x6 dx 1 x a Determine an appropriate trigonometric substitution Use x Use x sin 8 where tan 8 where T T 0 since the integrand contains the expression 1 x 2 2 dx T T 0 since the integrand contains the expression 1 x 2 2 Use x sec 0 where 0 0 or 0 since the integrand contains the expression 1 x V 2 3 2 b Apply the substitution to transform the integral into a trigonometric integral Do not evaluate the integral de

sin x Since cos x is the derivative of sin x then 12 sin x cos x dx can be done by substituting u Step 4 With the substitution u sin x we get 12 12 sin x cos x dx 12 u du 12 which integrates to C X Substituting back in to get the answer in terms of sin x we have S 12 sin x cos x dx C

Tutorial Exercise Evaluate the integral 1 Step 1 12 sin x cos x dx Since S 12 sin x cos x dx has an odd power of of cos x we will convert all but one power to sines We know that cos x sin x

Two marbles are drawn without replacement from a box with 4 blue 3 white and 2 red marbles Find the probability of selecting a blue and a red marble in order 0 111 O 0 486 O 0 591 0 227 0347

Consider the following 9 6 3 Let t 9 tan 0 and determine the following a dt t 0 0 dt 81 2 t 9 0 Evaluate the integral de

Tutorial Exercise Evaluate the integral Step 1 Since S 15 sin x cos x dx has an odd power of of cos x we will convert all but one power to sines We know that 15 sin x cos x dx cos x 1 11 sin x Step 2 Making this substitution using 1 15 sin x cos x dx gives us Step 3 Submit 15 15 sin x 1 sin x cos x dx 1 Skip you cannot come back 1 15 sin x cos x dx Since cos x is the derivative of sin x then 1 15 sin x cos x dx can be done by substituting u 15 sin x cos x 15 sin r cos r and du dx dx

Evaluate the integral I Step 1 15 sin x cos x dx Since 1 15 sin x cos x dx has an odd power of of cos x we will convert all but one power to sines We know that cos x 1 Step 2 Making this substitution using 1 gives us 1 sin x 15 sin x cos x dx 15 sin 15 sin x 1 sin x cos x dx 15 sin x cos x dx 155 SCC dx

Tutorial Exercise Evaluate the integral 15 Step 1 15 sin x cos x dx Since J 15 sin x cos x dx has an odd power of of cos x we will convert all but one power to sines We know that cos x sin x

Evaluate the integral Use C for the constant of integration 1 sin t cos t dt

Find the average value of f on 0 16 2 0 4 8 12 X

a Write an iterated integral for SS dA over the region bounded by y 7 x y 1 and x 0 using R vertical cross sections b Write an iterated integral for horizontal cross sections OA dx dy SS a Select the correct answer below and fill in the answer boxes to complete your choice OB S dy dx S SS dA over the region bounded by y 7 x y 1 and x 0 using R

Use Fubini s Theorem to evaluate Apply Fubini s Theorem 3xexy Gellan ng3 RECENTLY Honorer CROCHERISING SS3 00 exy dy dx Tyne exact opp 3xexy dx dy

1 Evaluate the following 2 a tan cos b sin cos Crat sin

bstitution integral Sx 2x 5 dx 28 5

Use Substitution to find the given indering integral Sx 2x 5 dx 3

find the given indefinite intergal state whether inte by Substitution or integrat integration by parts was used Please tell me if this parts or substitution dx use integration by L xex dx dx