7.3E: Exercises for Section 7.3

Simplify the expressions in exercises 1 - 5 by writing each one using a single trigonometric function.

1) \(4−4\sin^2θ\)

2) \(9\sec^2θ−9\)

Answer

\(9\sec^2θ−9 \quad = \quad 9\tan^2θ\)

3) \(a^2+a^2\tan^2θ\)

4) \(a^2+a^2\sinh^2θ\)

Answer

\(a^2+a^2\sinh^2θ \quad = \quad a^2\cosh^2θ\)

5) \(16\cosh^2θ−16\)

Use the technique of completing the square to express each trinomial in exercises 6 - 8 as the square of a binomial.

6) \(4x^2−4x+1\)

Answer

\( 4(x−\frac{1}{2})^2\)

7) \(2x^2−8x+3\)

8) \(−x^2−2x+4\)

Answer

\( −(x+1)^2+5\)

In exercises 9 - 28, integrate using the method of trigonometric substitution. Express the final answer in terms of the original variable.

9) \(\displaystyle ∫\frac{dx}{\sqrt{4−x^2}}\)

10) \(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}}\)

Answer

\(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}} \quad = \quad \ln∣x+\sqrt{−a^2+x^2}∣+C\)

11) \(\displaystyle ∫\sqrt{4−x^2}\,dx\)

12) \(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}}\)

Answer

\(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}} \quad = \quad \tfrac{1}{3}\ln∣\sqrt{9x^2+1}+3x∣+C\)

13) \(\displaystyle ∫\frac{x^2\,dx}{\sqrt{1−x^2}}\)

14) \(\displaystyle ∫\frac{dx}{x^2\sqrt{1−x^2}}\)

Answer

\(\displaystyle ∫\frac{dx}{x^2\sqrt{1−x^2}} \quad = \quad −\frac{\sqrt{1−x^2}}{x}+C\)

15) \(\displaystyle ∫\frac{dx}{(1+x^2)^2}\)

16) \(\displaystyle ∫\sqrt{x^2+9}\,dx\)

Answer

\(\displaystyle ∫\sqrt{x^2+9}\,dx \quad = \quad 9\left[\frac{x\sqrt{x^2+9}}{18}+\tfrac{1}{2}\ln\left|\frac{\sqrt{x^2+9}}{3}+\frac{x}{3}\right|\right]+C\)

17) \(\displaystyle ∫\frac{\sqrt{x^2−25}}{x}\,dx\)

18) \(\displaystyle ∫\frac{θ^3}{\sqrt{9−θ^2}}\,dθ\)

Answer

\(\displaystyle ∫\frac{θ^3dθ}{\sqrt{9−θ^2}}\,dθ \quad = \quad −\tfrac{1}{3}\sqrt{9−θ^2}(18+θ^2)+C\)

19) \(\displaystyle ∫\frac{dx}{\sqrt{x^6−x^2}}\)

20) \(\displaystyle ∫\sqrt{x^6−x^8}\,dx\)

Answer

\(\displaystyle ∫\sqrt{x^6−x^8}\,dx \quad = \quad \frac{(−1+x^2)(2+3x^2)\sqrt{x^6−x^8}}{15x^3}+C\)

21) \(\displaystyle ∫\frac{dx}{(1+x^2)^{3/2}}\)

22) \(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}}\)

Answer

\(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}} \quad = \quad −\frac{x}{9\sqrt{x^2-9}}+C\)

23) \(\displaystyle ∫\frac{\sqrt{1+x^2}}{x}\,dx\)

24) \(\displaystyle ∫\frac{x^2}{\sqrt{x^2−1}}\,dx\)

Answer

\(\displaystyle ∫\frac{x^2}{\sqrt{x^2−1}}\,dx \quad = \quad \tfrac{1}{2}(\ln∣x+\sqrt{x^2−1}∣+x\sqrt{x^2−1})+C\)

25) \(\displaystyle ∫\frac{x^2}{x^2+4}\,dx\)

26) \(\displaystyle ∫\frac{dx}{x^2\sqrt{x^2+1}}\)

Answer

\(\displaystyle ∫\frac{dx}{x^2\sqrt{x^2+1}} \quad = \quad −\frac{\sqrt{1+x^2}}{x}+C\)

27) \(\displaystyle ∫\frac{x^2}{\sqrt{1+x^2}}\,dx\)

28) \(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx\)

Answer

\(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx \quad = \quad \tfrac{1}{8}\left(x(5−2x^2)\sqrt{1−x^2}+3\arcsin x\right)+C\)

In exercises 29 - 34, use the substitutions \(x=\sinh θ\), \( \cosh θ\), or \(\tanh θ\). Express the final answers in terms of the variable \(x\).

29) \(\displaystyle ∫\frac{dx}{\sqrt{x^2−1}}\)

30) \(\displaystyle ∫\frac{dx}{x\sqrt{1−x^2}}\)

Answer

\(\displaystyle ∫\frac{dx}{x\sqrt{1−x^2}} \quad = \quad \ln x−\ln∣1+\sqrt{1−x^2}∣+C\)

31) \(\displaystyle ∫\sqrt{x^2−1}\,dx\)

32) \(\displaystyle ∫\frac{\sqrt{x^2−1}}{x^2}\,dx\)

Answer

\(\displaystyle ∫\frac{\sqrt{x^2−1}}{x^2}\,dx \quad = \quad −\frac{\sqrt{−1+x^2}}{x}+\ln\left|x+\sqrt{−1+x^2}\right|+C\)

33) \(\displaystyle ∫\frac{dx}{1−x^2}\)

34) \(\displaystyle ∫\frac{\sqrt{1+x^2}}{x^2}\,dx\)

Answer

\(\displaystyle ∫\frac{\sqrt{1+x^2}}{x^2}\,dx \quad = \quad −\frac{\sqrt{1+x^2}}{x}+\text{arcsinh}\, x+C\)

Use the technique of completing the square to evaluate the integrals in exercises 35 - 39.

35) \(\displaystyle ∫\frac{1}{x^2−6x}\,dx\)

36) \(\displaystyle ∫\frac{1}{x^2+2x+1}\,dx\)

Answer

\(\displaystyle ∫\frac{1}{x^2+2x+1}\,dx \quad = \quad −\frac{1}{1+x}+C\)

37) \(\displaystyle ∫\frac{1}{\sqrt{−x^2+2x+8}}\,dx\)

38) \(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx\)

Answer

\(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx \quad = \quad \arcsin\left( \frac{x-5}{5}\right)+C\)

39) \(\displaystyle ∫\frac{1}{\sqrt{x^2+4x−12}}\,dx\)

40) Evaluate the integral without using calculus: \(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx.\)

Answer

\(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx \quad = \quad \frac{9π}{2}\); area of a semicircle with radius 3

41) Find the area enclosed by the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1.\)

42) Evaluate the integral \(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}}\) using two different substitutions. First, let \(x=\cos θ\) and evaluate using trigonometric substitution. Second, let \(x=\sin θ\) and use trigonometric substitution. Are the answers the same?

Answer

\(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}} \quad = \quad \arcsin(x)+C\) is the common answer.

43) Evaluate the integral \(\displaystyle ∫\frac{dx}{x\sqrt{x^2−1}}\) using the substitution \(x=\sec θ\). Next, evaluate the same integral using the substitution \(x=\csc θ.\) Show that the results are equivalent.

44) Evaluate the integral \(\displaystyle ∫\frac{x}{x^2+1}\,dx\) using the form \(\displaystyle ∫\frac{1}{u}\,du\). Next, evaluate the same integral using \(x=\tan θ.\) Are the results the same?

Answer

\(\displaystyle ∫\frac{x}{x^2+1}\,dx \quad = \quad \frac{1}{2}\ln(1+x^2)+C\) is the result using either method.

45) State the method of integration you would use to evaluate the integral \(\displaystyle ∫x\sqrt{x^2+1}\,dx.\) Why did you choose this method?

46) State the method of integration you would use to evaluate the integral \(\displaystyle ∫x^2\sqrt{x^2−1}\,dx.\) Why did you choose this method?

Answer

Use trigonometric substitution. Let \(x=\sec(θ).\)

47) Evaluate \(\displaystyle ∫^1_{−1}\frac{x}{x^2+1}\,dx\)

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