# 5.5E: Exercises for Section 5.5

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1) Why is $$u$$-substitution referred to as a change of variable?

2) If $$f=g∘h$$, when reversing the chain rule, $$\dfrac{d}{dx}(g∘h)(x)=g′(h(x))h′(x)$$, should you take $$u=g(x)$$ or $$u=h(x)?$$

$$u=h(x)$$

In exercises 3 - 7, verify each identity using differentiation. Then, using the indicated $$u$$-substitution, identify $$f$$ such that the integral takes the form $$\displaystyle∫f(u)\,du.$$

3) $$\displaystyle ∫x\sqrt{x+1}\,dx=\frac{2}{15}(x+1)^{3/2}(3x−2)+C;\quad u=x+1$$

4) $$\displaystyle∫\frac{x^2}{\sqrt{x−1}}\,dx=\frac{2}{15}\sqrt{x−1}(3x^2+4x+8)+C,\quad (x>1);\quad u=x−1$$

$$f(u)=\dfrac{(u+1)^2}{\sqrt{u}}$$

5) $$\displaystyle∫x\sqrt{4x^2+9}\,dx=\frac{1}{12}(4x^2+9)^{3/2}+C;\quad u=4x^2+9$$

6) $$\displaystyle∫\frac{x}{\sqrt{4x^2+9}}\,dx=\frac{1}{4}\sqrt{4x^2+9}+C;\quad u=4x^2+9$$

$$du=8x\,dx;\quad f(u)=\frac{1}{8\sqrt{u}}$$

7) $$\displaystyle∫\frac{x}{(4x^2+9)^2}\,dx=−\frac{1}{8(4x^2+9)} + C;\quad u=4x^2+9$$

In exercises 8 - 17, find the antiderivative using the indicated substitution.

8) $$\displaystyle∫(x+1)^4\,dx;\quad u=x+1$$

$$\displaystyle∫(x+1)^4\,dx = \frac{1}{5}(x+1)^5+C$$

9) $$\displaystyle∫(x−1)^5\,dx;\quad u=x−1$$

10) $$\displaystyle∫(2x−3)^{−7}\,dx;\quad u=2x−3$$

$$\displaystyle∫(2x−3)^{−7}\,dx = −\frac{1}{12(2x−3)^6}+C$$

11) $$\displaystyle∫(3x−2)^{−11}\,dx;\quad u=3x−2$$

12) $$\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx;\quad u=x^2+1$$

$$\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx = \sqrt{x^2+1}+C$$

13) $$\displaystyle∫\frac{x}{\sqrt{1−x^2}}\,dx;\quad u=1−x^2$$

14) $$\displaystyle∫(x−1)(x^2−2x)^3\,dx;\quad u=x^2−2x$$

$$\displaystyle∫(x−1)(x^2−2x)^3\,dx = \frac{1}{8}(x^2−2x)^4+C$$

15) $$\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx;\quad u=x^3=3x^2$$

16) $$\displaystyle∫\cos^3 θ\,dθ;\quad u=\sin θ$$ (Hint: $$\cos^2 θ=1−\sin^2 θ$$)

$$\displaystyle∫\cos^3 θ\,dθ = \sin θ−\dfrac{\sin^3 θ}{3}+C$$

17) $$\displaystyle ∫\sin^3 θ\,dθ;\quad u=\cos θ$$ (Hint: $$\sin^2 θ=1−\cos^2θ$$)

In exercises 18 - 34, use a suitable change of variables to determine the indefinite integral.

18) $$\displaystyle∫x(1−x)^{99}\,dx$$

\begin{align*} \displaystyle∫x(1−x)^{99}\,dx &= \frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C \4pt] &=-\frac{(1-x)^{100}}{10100}\big[ 100x + 1 \big]+C \end{align*} 19) $$\displaystyle∫t(1−t^2)^{10}dt$$ 20) $$\displaystyle∫(11x−7)^{−3}\,dx$$ Answer $$\displaystyle∫(11x−7)^{−3}\,dx = −\frac{1}{22(11x−7)^2}+C$$ 21) $$\displaystyle∫(7x−11)^4\,dx$$ 22) $$\displaystyle∫\cos^3 θ\sin θ\,dθ$$ Answer $$\displaystyle∫\cos^3 θ\sin θ\,dθ = −\frac{\cos^4 θ}{4}+C$$ 23) $$\displaystyle∫\sin^7 θ\cos θ\,dθ$$ 24) $$\displaystyle∫\cos^2(πt)\sin(πt)\,dt$$ Answer $$\displaystyle∫\cos^2(πt)\sin(πt)\,dt = −\frac{cos^3(πt)}{3π}+C$$ 25) $$\displaystyle∫\sin^2 x\cos^3 x\,dx$$ (Hint: $$\sin^2 x+\cos^2 x=1$$) 26) $$\displaystyle∫t\sin(t^2)\cos(t^2)\,dt$$ Answer $$\displaystyle∫t\sin(t^2)\cos(t^2)\,dt = −\frac{1}{4}\cos^2(t^2)+C$$ 27) $$\displaystyle∫t^2\cos^2(t^3)\sin(t^3)\,dt$$ 28) $$\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx$$ Answer $$\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx = −\frac{1}{3(x^3−3)}+C$$ 29) $$\displaystyle∫\frac{x^3}{\sqrt{1−x^2}}\,dx$$ 30) $$\displaystyle∫\frac{y^5}{(1−y^3)^{3/2}}\,dy$$ Answer $$\displaystyle∫\frac{y^5}{(1−y^3)^{3/2}}\,dy = −\frac{2(y^3−2)}{3\sqrt{1−y^3}}+C$$ 31) $$\displaystyle∫\cos θ(1−\cos θ)^{99}\sin θ\,dθ$$ 32) $$\displaystyle∫(1−\cos^3 θ)^{10}\cos^2 θ\sin θ\,dθ$$ Answer $$\displaystyle∫(1−\cos^3 θ)^{10}\cos^2 θ\sin θ\,dθ = \frac{1}{33}(1−\cos^3 θ)^{11}+C$$ 33) $$\displaystyle∫(\cos θ−1)(\cos^2 θ−2\cos θ)^3\sin θ\,dθ$$ 34) $$\displaystyle∫(\sin^2 θ−2\sin θ)(\sin^3 θ−3\sin^2 θ)^3\cos θ\,dθ$$ Answer $$\displaystyle∫(\sin^2 θ−2\sin θ)(\sin^3 θ−3\sin^2 θ)^3\cos θ\,dθ = \frac{1}{12}(\sin^3 θ−3\sin^2 θ)^4+C$$ In exercises 35 - 38, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer. 35) [T] $$y=3(1−x)^2$$ over $$[0,2]$$ 36) [T] $$y=x(1−x^2)^3$$ over $$[−1,2]$$ Answer $$L_{50}=−8.5779.$$ The exact area is $$\frac{−81}{8}$$ units$$^2$$. 37) [T] $$y=\sin x(1−\cos x)^2$$ over $$[0,π]$$ 38) [T] $$y=\dfrac{x}{(x^2+1)^2}$$ over $$[−1,1]$$ Answer $$L_{50}=−0.006399$$. The exact area is 0. In exercises 39 - 44, use a change of variables to evaluate the definite integral. 39) $$\displaystyle∫^1_0x\sqrt{1−x^2}\,dx$$ 40) $$\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx$$ Answer $$\displaystyle u=1+x^2,\quad du=2x\,dx,\quad ∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1$$ 41) $$\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}\,dt$$ 42) $$\displaystyle∫^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt$$ Answer $$\displaystyle u=1+t^3,\quad du=3t^2,\quad ∫^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt = \frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)$$ 43) $$\displaystyle∫^{π/4}_0\sec^2 θ\tan θ\,dθ$$ 44) $$\displaystyle∫^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ$$ Answer $$\displaystyle u=\cos θ,\quad du=−\sin θ\,dθ,\quad \int^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ = -∫_1^{\sqrt{2}/2}u^{−4}\,du = ∫^1_{\sqrt{2}/2}u^{−4}\,du=\frac{1}{3}(2\sqrt{2}−1)$$ In exercises 45 - 50, evaluate the indefinite integral $$\displaystyle ∫f(x)\,dx$$ with constant $$C=0$$ using $$u$$-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of $$C$$ that would need to be added to the antiderivative to make it equal to the definite integral $$\displaystyle F(x)=∫^x_af(t)\,dt$$, with a the left endpoint of the given interval. 45) [T] $$\displaystyle∫(2x+1)e^{x^2+x−6}\,dx$$ over $$[−3,2]$$ 46) [T] $$\displaystyle∫\frac{\cos(\ln(2x))}{x}\,dx$$ on $$[0,2]$$ Answer The antiderivative is $$y=\sin(\ln(2x))$$. Since the antiderivative is not continuous at $$x=0$$, one cannot find a value of C that would make $$y=\sin(\ln(2x))−C$$ work as a definite integral. 47) [T] $$\displaystyle ∫\frac{3x^2+2x+1}{\sqrt{x^3+x^2+x+4}}\,dx$$ over $$[−1,2]$$ 48) [T] $$\displaystyle ∫\frac{\sin x}{\cos^3x}\,dx$$ over $$\left[−\frac{π}{3},\frac{π}{3}\right]$$ Answer The antiderivative is $$y=\frac{1}{2}\sec^2 x$$. You should take $$C=−2$$ so that $$F(−\frac{π}{3})=0.$$ 49) [T] $$\displaystyle ∫(x+2)e^{−x^2−4x+3}\,dx$$ over $$[−5,1]$$ 50) [T] $$\displaystyle ∫3x^2\sqrt{2x^3+1}\,dx$$ over $$[0,1]$$ Answer The antiderivative is $$y=\frac{1}{3}(2x^3+1)^{3/2}$$. One should take $$C=−\frac{1}{3}$$. 51) If $$h(a)=h(b)$$ in $$\displaystyle ∫^b_ag'(h(x))h(x)\,dx,$$ what can you say about the value of the integral? 52) Is the substitution $$u=1−x^2$$ in the definite integral $$\displaystyle ∫^2_0\frac{x}{1−x^2}\,dx$$ okay? If not, why not? Answer No, because the integrand is discontinuous at $$x=1$$. In exercises 53 - 59, use a change of variables to show that each definite integral is equal to zero. 53) $$\displaystyle ∫^π_0\cos^2(2θ)\sin(2θ)\,dθ$$ 54) $$\displaystyle ∫^\sqrt{π}_0t\cos(t^2)\sin(t^2)\,dt$$ Answer $$u=\sin(t^2);$$ the integral becomes $$\displaystyle \frac{1}{2}∫^0_0u\,du.$$ 55) $$\displaystyle ∫^1_0(1−2t)\,dt$$ 56) $$\displaystyle ∫^1_0\frac{1−2t}{1+(t−\frac{1}{2})^2}\,dt$$ Answer $$u=1+(t−\frac{1}{2})^2;$$ the integral becomes $$\displaystyle −∫^{5/4}_{5/4}\frac{1}{u}\,du$$. 57) $$\displaystyle ∫^π_0\sin\left(\left(t−\tfrac{π}{2}\right)^3\right)\cos\left(t−\tfrac{π}{2}\right)\,dt$$ 58) $$\displaystyle ∫^2_0(1−t)\cos(πt)\,dt$$ Answer $$u=1−t;$$ Since the integrand is odd, the integral becomes \[∫^{−1}_1u\cos\big(π(1−u)\big)\,du=∫^{−1}_1u[\cos π\cos u−\sin π\sin u]\,du=−∫^{−1}_1u\cos u\,du=∫_{-1}^1u\cos u\,du=0\nonumber

59) $$\displaystyle ∫^{3π/4}_{π/4}\sin^2 t\cos t\,dt$$

60) Show that the average value of $$f(x)$$ over an interval $$[a,b]$$ is the same as the average value of $$f(cx)$$ over the interval $$\left[\frac{a}{c},\frac{b}{c}\right]$$ for $$c>0.$$

Setting $$u=cx$$ and $$du=c\,dx$$ gets you $$\displaystyle \frac{1}{\frac{b}{c}−\frac{a}{c}}∫^{b/c}_{a/c}f(cx)\,dx=\frac{c}{b−a}∫^{u=b}_{u=a}f(u)\frac{du}{c}=\frac{1}{b−a}∫^b_af(u)\,du.$$

61) Find the area under the graph of $$f(t)=\dfrac{t}{(1+t^2)^a}$$ between $$t=0$$ and $$t=x$$ where $$a>0$$ and $$a≠1$$ is fixed, and evaluate the limit as $$x→∞$$.

62) Find the area under the graph of $$g(t)=\dfrac{t}{(1−t^2)^a}$$ between $$t=0$$ and $$t=x$$, where $$0<x<1$$ and $$a>0$$ is fixed. Evaluate the limit as $$x→1$$.

$$\displaystyle ∫^x_0g(t)\,dt=\frac{1}{2}∫^1_{u=1−x^2} \frac{du}{u^a}=\frac{1}{2(1−a)}u^{1−a}∣1u=\frac{1}{2(1−a)}(1−(1−x^2)^{1−a})$$ As $$x→1$$ the limit is $$\dfrac{1}{2(1−a)}$$ if $$a<1$$, and the limit diverges to $$+∞$$ if $$a>1$$.

63) The area of a semicircle of radius $$1$$ can be expressed as $$\displaystyle ∫^1_{−1}\sqrt{1−x^2}\,dx$$. Use the substitution $$x=\cos t$$ to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

64) The area of the top half of an ellipse with a major axis that is the $$x$$-axis from $$x=−1$$ to a and with a minor axis that is the $$y$$-axis from $$y=−b$$ to $$y=b$$ can be written as $$\displaystyle ∫^a_{−a}b\sqrt{1−\frac{x^2}{a^2}}\,dx$$. Use the substitution $$x=a\cos t$$ to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

$$\displaystyle ∫^{t=0}_{t=π}b\sqrt{1−\cos^2 t}×(−a\sin t)\,dt=∫^{t=π}_{t=0}ab\sin^2 t\,dt$$

65) [T] The following graph is of a function of the form $$f(t)=a\sin(nt)+b\sin(mt)$$. Estimate the coefficients $$a$$ and $$b$$ and the frequency parameters $$n$$ and $$m$$. Use these estimates to approximate $$\displaystyle ∫^π_0f(t)\,dt$$. 66) [T] The following graph is of a function of the form $$f(x)=a\cos(nt)+b\cos(mt)$$. Estimate the coefficients $$a$$ and $$b$$ and the frequency parameters $$n$$ and $$m$$. Use these estimates to approximate $$\displaystyle ∫^π_0f(t)\,dt.$$ $$f(t)=2\cos(3t)−\cos(2t);\quad \displaystyle ∫^{π/2}_0(2\cos(3t)−\cos(2t))\,dt=−\frac{2}{3}$$