Chapter 9 Solutions

Authored worked explanation for the original source problem.

1. Let u=csc x+cot x. Then $u\prime =-\csc x\cot x-{\csc }^{2}x=-\csc x·u$.

   Why: The derivative combines into the same selected sum.

2. Multiplying the integrand by u/u gives $\int \csc x·u/udx=-\int du/u$.

   Why: The substitution supplies the numerator and leaves the log denominator.

3. Integrate to −ln|u|+C and substitute back.

   Why: Absolute values cover any connected interval where sine is nonzero.

Reviewed answer: $-\ln |\csc x+\cot x|+C$, on intervals with sin x≠0.

Takeaway: A well-chosen sum can make a reciprocal trig integral logarithmic.

Authored worked explanation for the original source problem.

1. A primitive is $F\left(x\right)=\ln \left|\sec \right(\sin x)+\tan (\sin x\left)\right|$.

   Why: Differentiating gives sec(sin x)·cos x, equal to the integrand.

2. The inner sine is 0 at both x=0 and x=π, so F has equal endpoint values.

   Why: The integrand is continuous because sin x lies in [0,1], where cosine of that inner value stays positive.

3. The definite integral is F(π)−F(0)=0.

   Why: This is cancellation of signed contributions, not zero geometric area.

Reviewed answer: $0$

Takeaway: Using a primitive avoids treating a nonmonotone substitution as a one-to-one interval change.

Authored worked explanation for the original source problem.

Source review note: The source gives 36.3. The printed MathML has upper bound 2 followed by coefficient 6; evaluating that integral yields the reviewed value shown here.

1. The printed upper limit is 2 and the coefficient of the radical is 6. Set u=√x, so dx=2u du and u runs from 0 to √2.

   Why: Read the MathML bounds separately from the adjacent coefficient.

2. The integral becomes $12{{\int }_0}^{\sqrt{2}}u\sqrt{4-u}du$. Put v=4−u to integrate $-12(4-v)v^{1/2}dv$.

   Why: The second substitution removes the radical's shift.

3. A primitive in x is $-32(4-\sqrt{x}{)}^{3/2}+(24/5\left)\right(4-\sqrt{x}{)}^{5/2}$; subtract its value at 0 from its value at 2.

   Why: The lower primitive value is −512/5, so its subtraction adds 512/5.

Reviewed answer: $512/5-32(4-\sqrt{2}{)}^{3/2}+(24/5\left)\right(4-\sqrt{2}{)}^{5/2}\approx 20.951317$.

Takeaway: Nested substitutions require both differentials and careful endpoint readings.

Authored worked explanation for the original source problem.

1. Use x=3sinθ with θ in [−π/2,π/2]; dx=3cosθ dθ and the radical is 3cosθ.

   Why: This angle range makes the square root's cosine nonnegative.

2. The transformed integral is $6\int {\cos }^2\theta d\theta =3\theta +(3/2)\sin \left(2\theta \right)+C$.

   Why: Use cos²θ=(1+cos2θ)/2.

3. Since sin2θ=2(x/3)(√(9−x²)/3), substitute back to the displayed expression.

   Why: This form is equivalent to the source's trig expression on the interior real interval.

Reviewed answer: $(x/3)\sqrt{9-x^2}+3\arcsin (x/3)+C$

Takeaway: The sign choice for the radical belongs to the substitution's angle interval.

Authored worked explanation for the original source problem.

Source review note: The source gives ln x−2x²+C, valid only for x>0. The original integrand also has a real branch on x<0, covered by ln|x|.

1. Split the integrand into 1/x−4x.

   Why: Divide the numerator terms separately.

2. Integrate the two parts to ln|x|−2x²+C.

   Why: The absolute-value logarithm includes both nonzero sign intervals.

Reviewed answer: $\ln \left|x\right|-2x^2+C$, x≠0.

Takeaway: A primitive's domain should match the original rational function.

Authored worked explanation for the original source problem.

Source review note: The source value is consistent after cancelling a removable 0/0 at x=2 and evaluating the integral by limiting/continuous-extension interpretation.

1. Factor the denominator as $(2-x)(x^2+4)$; away from x=2 the quotient equals 1/(x²+4).

   Why: The original 0/0 at 2 remains a source-domain hole.

2. The simplified function has a continuous extension through 2 with value 1/8. The two improper side limits at that point are finite.

   Why: A single removable hole does not change the definite integral when interpreted via limits.

3. A primitive is (1/2)arctan(x/2); evaluating from 0 to infinity gives (1/2)(π/2)=π/4.

   Why: Also take the infinite endpoint as a limit.

Reviewed answer: $\pi /4$, treating x=2 as a removable hole.

Takeaway: A removable hole and a genuine singularity behave differently in integration.

Authored worked explanation for the original source problem.

1. Because x>0 on [1,3], the integrand simplifies to $\sqrt{2x-x^2+3}=\sqrt{4-(x-1{)}^2}$.

   Why: Combining the positive square roots does not introduce an absolute-value sign error.

2. Let u=x−1, giving ${{\int }_0}^2\sqrt{4-u^2}du$.

   Why: The shifted radical describes a quarter circle of radius 2.

3. Its area is $(1/4)\pi ·2^2=\pi$; a trig substitution u=2sinθ gives the same value.

   Why: The integration interval stays within the real radical domain.

Reviewed answer: $\pi$

Takeaway: Geometry can confirm an otherwise long trig substitution.

Authored worked explanation for the original source problem.

Source review note: The source gives cos(ln(cos x))+3+C. The denominator requires a logarithm of that expression.

1. Choose $u=\cos \left(\ln \right(\cos x\left)\right)+3$.

   Why: The denominator is the natural candidate if its derivative appears above.

2. $u\prime =-\sin \left(\ln \right(\cos x\left)\right)·(-\tan x)=\tan x\sin \left(\ln \right(\cos x\left)\right)$.

   Why: The two chain-rule minus signs cancel.

3. The integral is ∫du/u=ln u+C; u≥2 so no absolute value is needed.

   Why: The inside ln requires cos x>0, and the outer log argument stays positive.

Reviewed answer: $\ln \left(\cos \right(\ln \left(\cos x\right))+3)+C$, on cos x>0.

Takeaway: A denominator paired with its derivative gives a logarithm.

Authored worked explanation for the original source problem.

Source review note: The source gives ln(ln(cos(ln x)))+3+C. The printed denominator includes +3 inside its bracket; its primitive is −ln|ln(cos(ln x))+3|+C, and the source expression is generally not real-defined.

1. Factor the denominator as $x\left[\ln \right(\cos \left(\ln x\right))+3]$ and set $u=\ln \left(\cos \right(\ln x\left)\right)+3$.

   Why: The whole bracket is the repeated inside expression.

2. $u\prime =-\tan \left(\ln x\right)/x$.

   Why: Differentiate the outer logarithm, cosine, and inner logarithm in order.

3. The integral becomes −∫du/u, giving the displayed answer and its domain restrictions.

   Why: The negative differential supplies the sign; the absolute value allows either nonzero sign of u.

Reviewed answer: $-\ln \left|\ln \right(\cos \left(\ln x\right))+3|+C$ where x>0, cos(ln x)>0, and ln(cos(ln x))+3≠0.

Takeaway: A missing minus sign or misplaced +3 changes a nested primitive.

Authored worked explanation for the original source problem.

1. Integrate the constant 1 to x. For the remaining term use u=1−x², du=−2x dx.

   Why: Split the sum and match the radical's inside derivative.

2. The substituted part is ∫u^(1/2)du=(2/3)u^(3/2).

   Why: The negative coefficient was already absorbed in du.

3. Return to x and add C.

   Why: The radical restricts the original real interval.

Reviewed answer: $x+(2/3)(1-x^2{)}^{3/2}+C$ on −1<x<1.

Takeaway: One term can be elementary while another needs substitution.

Authored solution to an added practice problem.

1. Let u=x²+1, du=2x dx.

   Why: The numerator exactly matches the inside derivative.

2. The integral is ∫u³du=u⁴/4+C, then substitute back.

   Why: After changing variables, no x remains.

Reviewed answer: $(x^2+1{)}^4/4+C$

Takeaway: A complete substitution changes both the function and differential.

Authored solution to an added practice problem.

1. Set u=x²+4, with du=2x dx.

   Why: The denominator's derivative appears in the numerator.

2. Integrate du/u and return to x; x²+4>0 for all real x.

   Why: A positive argument makes the absolute value unnecessary.

Reviewed answer: $\ln (x^2+4)+C$

Takeaway: A positive quadratic gives a globally valid log argument.

Authored solution to an added practice problem.

1. Let u=x³, so du=3x² dx.

   Why: The coefficient mismatch leaves a factor 1/3.

2. The integral becomes (1/3)∫sin u du=−cos u/3+C.

   Why: The primitive of sine has a minus sign.

Reviewed answer: $-(1/3)\cos \left(x^3\right)+C$

Takeaway: Carry a differential's constant factor through the integral.

Authored solution to an added practice problem.

1. Set u=e^x+2, giving du=e^x dx.

   Why: The numerator supplies the inside derivative.

2. The result is ln u+C and u>0 for every real x.

   Why: The reciprocal u pattern has a logarithmic primitive.

Reviewed answer: $\ln (e^x+2)+C$

Takeaway: A constant shift does not change an exponential's derivative.

Authored solution to an added practice problem.

1. Let u=ln x, so du=dx/x.

   Why: The denominator x supplies the log differential.

2. Integrate u² to u³/3+C and return to x.

   Why: The substitution accounts for the entire integrand.

Reviewed answer: $(\ln x{)}^3/3+C$ for x>0.

Takeaway: A logarithm often pairs with a reciprocal x factor.

Authored solution to an added practice problem.

1. Use u=x²+1 and du=2x dx; bounds x=0,2 become u=1,5.

   Why: Transform each endpoint through the same substitution.

2. The integral is ∫₁⁵du/u=ln5−ln1=ln5.

   Why: Evaluate in the variable used by the differential.

Reviewed answer: $\ln 5$

Takeaway: Changed bounds prevent mixing x and u in the final evaluation.

Authored solution to an added practice problem.

1. Let u=1−x², du=−2x dx; x=0,1 become u=1,0.

   Why: The substitution reverses the endpoint order.

2. The integral becomes $-(1/2){{\int }_1}^0{u}^{-1/2}du=(1/2){{\int }_0}^1{u}^{-1/2}du=1$.

   Why: Reversing the bounds cancels the differential's negative sign.

Reviewed answer: $1$

Takeaway: Bounds and differential signs must be tracked together.

Authored solution to an added practice problem.

1. Divide: $(x^2+1)/(x-1)=x+1+2/(x-1)$.

   Why: A higher-degree numerator can be simplified before selecting a substitution.

2. Integrate the polynomial terms and use u=x−1 for the reciprocal term.

   Why: The linear shift's differential is dx.

Reviewed answer: $x^2/2+x+2\ln |x-1|+C$, x≠1.

Takeaway: Algebra can be the most useful first integration step.

Authored solution to an added practice problem.

1. Use x=2sinθ for 0≤θ≤π/6; dx=2cosθ dθ and (4−x²)^(3/2)=8cos³θ.

   Why: The cosine is nonnegative on this chosen interval.

2. The integral becomes $(1/4){{\int }_0}^{\pi /6}{\sec }^2\theta d\theta =(1/4)\tan (\pi /6)$.

   Why: The differential’s cosine cancels one factor from the denominator’s cos³θ, leaving 1/(4cos²θ).

Reviewed answer: $1/\left[4\sqrt{3}\right]$

Takeaway: Include the scale factor from both the radical and differential.

Authored solution to an added practice problem.

1. The graph y=√(16−x²) on [0,4] is a quarter circle of radius 4.

   Why: The radicand is nonnegative through both endpoints.

2. Its area is π·4²/4=4π. Equivalently x=4sinθ yields $16{{\int }_0}^{\pi /2}{\cos }^2\theta d\theta$.

   Why: Geometry confirms the trig substitution and its bounds.

Reviewed answer: $4\pi$

Takeaway: A geometric check can catch an omitted scale factor.

Authored solution to an added practice problem.

1. Complete the square: $2x-x^2=1-(x-1{)}^2$.

   Why: A shift turns the radical into a standard circle.

2. For x from 1 to 2, u=x−1 runs from 0 to 1; the integral is one quarter of a unit circle.

   Why: The transformed interval selects the right-hand quarter.

Reviewed answer: $\pi /4$

Takeaway: Complete the square before choosing a trig substitution.

Authored solution to an added practice problem.

1. Let x=3tanθ, dx=3sec²θ dθ and x²+9=9sec²θ.

   Why: The tangent identity handles a sum of squares.

2. Bounds x=0,3 become θ=0,π/4, so the integral is $(1/3){{\int }_0}^{\pi /4}d\theta =\pi /12$.

   Why: The secant squared factors cancel exactly.

Reviewed answer: $\pi /12$

Takeaway: Choose tan for a positive sum of squares.

Authored solution to an added practice problem.

1. Let u=sec x+tan x, so du=sec x(sec x+tan x)dx.

   Why: The chosen sum factors from its derivative.

2. Multiply sec x by u/u and obtain ∫du/u.

   Why: This leaves a logarithmic primitive.

3. Restore x and add C.

   Why: The source function's excluded cosine zeros remain excluded.

Reviewed answer: $\ln |\sec x+\tan x|+C$, cos x≠0.

Takeaway: A clever factor of one can create the needed differential.

Authored solution to an added practice problem.

1. Rewrite tan x as sin x/cos x and let u=cos x.

   Why: The denominator's derivative is almost the numerator.

2. du=−sin x dx, so the integral is −∫du/u.

   Why: The differential's sign gives the leading minus.

Reviewed answer: $-\ln \left|\cos x\right|+C$, cos x≠0.

Takeaway: A trig quotient can be an ordinary logarithmic substitution.

Authored solution to an added practice problem.

1. Use ${\sin }^2\left(2x\right)=(1-\cos (4x\left)\right)/2$.

   Why: The half-angle identity lowers the squared trig power.

2. Integrate the constant 1/2 and the scaled cosine separately.

   Why: The cosine argument 4x contributes a division by 4.

Reviewed answer: $x/2-\sin \left(4x\right)/8+C$

Takeaway: An identity can make an integral elementary before substitution.

Authored solution to an added practice problem.

1. For u=x², du=2x dx, so x dx=du/2.

   Why: The original numerator contains x, not 2x.

2. The proposed ∫e^u du is twice the correct result.

   Why: Every differential's coefficient must be accounted for.

Reviewed answer: The correct rewritten integral is $(1/2)\int e^{u}du$, hence $(1/2)e^{x^2}+C$.

Takeaway: Write the differential explicitly before replacing the integrand.

Authored solution to an added practice problem.

1. In general $\sqrt{{\cos }^2\theta }=\left|\cos \theta \right|$.

   Why: A square root returns the nonnegative value.

2. Choosing θ in [−π/2,π/2] ensures cosθ≥0, so the absolute value may then be removed.

   Why: A substitution's angle interval justifies its simplified radical.

Reviewed answer: Use $\sqrt{1-{\sin }^2\theta }=\cos \theta$ on θ∈[−π/2,π/2]; outside that interval the absolute value is needed.

Takeaway: A branch choice is part of a trig substitution, not an optional afterthought.

Authored solution to an added practice problem.

1. Set u=3+sin x, du=cos x dx.

   Why: The denominator and its derivative pair exactly.

2. The log needs no absolute value because u lies in [2,4].

   Why: Its argument stays positive on the entire real line.

Reviewed answer: $\ln (3+\sin x)+C$

Takeaway: Check the sign of a log argument to simplify the final form safely.

Authored solution to an added practice problem.

1. A primitive of $2x\cos \left(x^2\right)$ is sin(x²).

   Why: The inner square's derivative is the numerator.

2. At −1 and 1 the primitive is the same, sin1, so the difference is zero.

   Why: The odd integrand's negative and positive portions also cancel.

Reviewed answer: $0$

Takeaway: Equal transformed endpoint values can signal signed cancellation.

Authored solution to an added practice problem.

1. For u=4−x², du=−2x dx, but the original integral has no factor x.

   Why: The straightforward substitution does not absorb the whole differential.

2. With x=2sinθ on the principal sine interval, √(4−x²)=2cosθ and dx=2cosθ dθ.

   Why: The trig identity removes the radical and accounts for the differential together.

3. The remaining cos²θ can be integrated by a half-angle identity.

   Why: The transformed expression uses familiar trig integrals.

Reviewed answer: Use x=2sinθ, which transforms the integrand to 4cos²θ dθ; u=4−x² alone leaves an unmatched x factor.

Takeaway: Choose the substitution that simplifies the full integral, including dx.