
# 5.7E: Exercises for Integrals Resulting in Inverse Trigonometric Functions


In exercises 1 - 6, evaluate each integral in terms of an inverse trigonometric function.

1) $$\displaystyle ∫^{\sqrt{3}/2}_0\frac{dx}{\sqrt{1−x^2}}$$

$$\displaystyle ∫^{\sqrt{3}/2}_0\frac{dx}{\sqrt{1−x^2}} \quad = \quad \arcsin x\bigg|^{\sqrt{3}/2}_0=\dfrac{π}{3}$$

2) $$\displaystyle ∫^{1/2}_{−1/2}\frac{dx}{\sqrt{1−x^2}}$$

3) $$\displaystyle ∫^1_{\sqrt{3}}\frac{dx}{1+x^2}$$

$$\displaystyle ∫^1_{\sqrt{3}}\frac{dx}{1+x^2} \quad = \quad \arctan x\bigg|^1_{\sqrt{3}}=−\dfrac{π}{12}$$

4) $$\displaystyle ∫^{\sqrt{3}}_{\frac{1}{\sqrt{3}}}\frac{dx}{1+x^2}$$

5) $$\displaystyle ∫^{\sqrt{2}}_1\frac{dx}{|x|\sqrt{x^2−1}}$$

$$\displaystyle ∫^{\sqrt{2}}_1\frac{dx}{|x|\sqrt{x^2−1}} \quad = \quad \text{arcsec}\, x\bigg|^{\sqrt{2}}_1=\dfrac{π}{4}$$

6) $$\displaystyle ∫^{\frac{2}{\sqrt{3}}}_1\frac{dx}{|x|\sqrt{x^2−1}}$$

In exercises 7 - 12, find each indefinite integral, using appropriate substitutions.

7) $$\displaystyle ∫\frac{dx}{\sqrt{9−x^2}}$$

$$\displaystyle ∫\frac{dx}{\sqrt{9−x^2}} \quad = \quad \arcsin \left(\frac{x}{3}\right)+C$$

8) $$\displaystyle ∫\frac{dx}{\sqrt{1−16x^2}}$$

$$\displaystyle ∫\frac{dx}{\sqrt{1−16x^2}} = \frac{1}{4}∫\frac{4\,dx}{\sqrt{1−(4x)^2}} \quad = \quad \frac{1}{4}\arcsin \left(4x\right)+C$$

9) $$\displaystyle ∫\frac{dx}{9+x^2}$$

$$\displaystyle ∫\frac{dx}{9+x^2} \quad = \quad \frac{1}{3}\arctan \left(\frac{x}{3}\right)+C$$

10) $$\displaystyle ∫\frac{dx}{25+16x^2}$$

$$\displaystyle ∫\frac{dx}{25+16x^2} = ∫\frac{dx}{5^2+(4x)^2} = \frac{1}{4}∫\frac{4\,dx}{5^2+(4x)^2} = \frac{1}{4}\cdot\frac{1}{5}\arctan \left(\frac{4x}{5}\right)+C\quad = \quad \frac{1}{20}\arctan \left(\frac{4x}{5}\right)+C$$

11) $$\displaystyle ∫\frac{dx}{x\sqrt{x^2−9}}$$

$$\displaystyle ∫\frac{dx}{x\sqrt{x^2−9}} \quad = \quad \frac{1}{3}\text{arcsec} \left(\frac{|x|}{3}\right)+C$$

12) $$\displaystyle ∫\frac{dx}{x\sqrt{4x^2−16}}$$

$$\displaystyle ∫\frac{dx}{x\sqrt{4x^2−16}} = ∫\frac{2\,dx}{2x\sqrt{(2x)^2−4^2}} = ∫\frac{du}{u\sqrt{u^2−4^2}} = \frac{1}{4}\text{arcsec} \left(\frac{|u|}{4}\right)+C \quad = \quad \frac{1}{4}\text{arcsec} \left(\frac{|x|}{2}\right)+C$$

13) Explain the relationship $$\displaystyle −\arccos t+C=∫\frac{dt}{\sqrt{1−t^2}}=\arcsin t+C.$$ Is it true, in general, that $$\arccos t=−\arcsin t$$?

$$\cos(\frac{π}{2}−θ)=\sin θ.$$ So, $$\arcsin t=\dfrac{π}{2}−\arccos t.$$ They differ by a constant.

14) Explain the relationship $$\displaystyle \text{arcsec}\, t+C=∫\frac{dt}{|t|\sqrt{t^2−1}}=−\text{arccsc}\, t+C.$$ Is it true, in general, that $$\text{arcsec}\, t=−\text{arccsc}\, t$$?

15) Explain what is wrong with the following integral: $$\displaystyle ∫^2_1\frac{dt}{\sqrt{1−t^2}}$$.

$$\sqrt{1−t^2}$$ is not defined as a real number when $$t>1$$.

16) Explain what is wrong with the following integral: $$\displaystyle ∫^1_{−1}\frac{dt}{|t|\sqrt{t^2−1}}$$.

$$\sqrt{t^2−1}$$ is not defined as a real number when $$-1 \lt t \lt 1$$, and the integrand is undefined when $$t = -1$$ or $$t = 1$$.

In exercises 17 - 20, solve for the antiderivative of $$f$$ with $$C=0$$, then use a calculator to graph $$f$$ and the antiderivative over the given interval $$[a,b]$$. Identify a value of $$C$$ such that adding $$C$$ to the antiderivative recovers the definite integral $$\displaystyle F(x)=∫^x_af(t)\,dt$$.

17) [T] $$\displaystyle ∫\frac{1}{\sqrt{9−x^2}}\,dx$$ over $$[−3,3]$$

The antiderivative is $$\arcsin (\frac{x}{3})+C$$. Taking $$C=\frac{π}{2}$$ recovers the definite integral.

18) [T] $$\displaystyle ∫\frac{9}{9+x^2}\,dx$$ over $$[−6,6]$$

19) [T] $$\displaystyle ∫\frac{\cos x}{4+\sin^2x}\,dx$$ over $$[−6,6]$$

The antiderivative is $$\frac{1}{2}\arctan (\frac{\sin x}{2})+C$$. Taking $$C=\frac{1}{2}\arctan (\frac{\sin(6)}{2})$$ recovers the definite integral.

20) [T] $$\displaystyle ∫\frac{e^x}{1+e^{2x}}\,dx$$ over $$[−6,6]$$

In exercises 21 - 26, compute the antiderivative using appropriate substitutions.

21) $$\displaystyle ∫\frac{\arcsin t}{\sqrt{1−t^2}}\,dt$$

$$\displaystyle ∫\frac{\arcsin t\,dt}{\sqrt{1−t^2}} \quad = \quad \tfrac{1}{2}(\arcsin t)^2+C$$

22) $$\displaystyle ∫\frac{dt}{\arcsin t\sqrt{1−t^2}}$$

$$\displaystyle ∫\frac{dt}{\arcsin t\sqrt{1−t^2}} \quad = \quad \ln\left|\arcsin t\right|+C$$

23) $$\displaystyle ∫\frac{\arctan (2t)}{1+4t^2}\,dt$$

$$\displaystyle ∫\frac{\arctan (2t)}{1+4t^2}\,dt \quad = \quad \frac{1}{4}(\arctan (2t))^2+C$$

24) $$\displaystyle ∫\frac{t\arctan (t^2)}{1+t^4}\,dt$$

$$\displaystyle ∫\frac{t\arctan (t^2)}{1+t^4}\,dt = \frac{1}{2}∫u\,du = \frac{1}{2}\cdot\frac{u^2}{2}+C\quad = \quad \frac{\left(\arctan \left(t^2\right)\right)^2}{4}+C$$

25) $$\displaystyle ∫\frac{\text{arcsec} \left(\tfrac{t}{2}\right)}{|t|\sqrt{t^2−4}}\,dt$$

$$\displaystyle ∫\frac{\text{arcsec} \left(\tfrac{t}{2}\right)}{|t|\sqrt{t^2−4}}\,dt \quad = \quad \tfrac{1}{4}(\text{arcsec} \left(\tfrac{t}{2}\right))^2+C$$

26) $$\displaystyle ∫\frac{t\,\text{arcsec} (t^2)}{t^2\sqrt{t^4−1}}\,dt$$

In exercises 27 - 32, use a calculator to graph the antiderivative of $$f$$ with $$C=0$$ over the given interval $$[a,b].$$ Approximate a value of $$C$$, if possible, such that adding $$C$$ to the antiderivative gives the same value as the definite integral $$\displaystyle F(x)=∫^x_af(t)\,dt.$$

27) [T] $$\displaystyle ∫\frac{1}{x\sqrt{x^2−4}}\,dx$$ over $$[2,6]$$

The antiderivative is $$\frac{1}{2}\text{arcsec} (\frac{x}{2})+C$$. Taking $$C=0$$ recovers the definite integral over $$[2,6]$$.

28) [T] $$\displaystyle ∫\frac{1}{(2x+2)\sqrt{x}}\,dx$$ over $$[0,6]$$

29) [T] $$\displaystyle ∫\frac{(\sin x+x\cos x)}{1+x^2\sin^2x\,dx}$$ over $$[−6,6]$$

The general antiderivative is $$\arctan (x\sin x)+C$$. Taking $$C=−\arctan (6\sin(6))$$ recovers the definite integral.

30) [T] $$\displaystyle ∫\frac{2e^{−2x}}{\sqrt{1−e^{−4x}}}\,dx$$ over $$[0,2]$$

31) [T] $$\displaystyle ∫\frac{1}{x+x\ln 2x}$$ over $$[0,2]$$

The general antiderivative is $$\arctan (\ln x)+C$$. Taking $$\displaystyle C=\tfrac{π}{2}=\lim_{t \to ∞}\arctan t$$ recovers the definite integral.

32) [T] $$\displaystyle ∫\frac{\arcsin x}{\sqrt{1−x^2}}$$ over $$[−1,1]$$

In exercises 33 - 38, compute each integral using appropriate substitutions.

33) $$\displaystyle ∫\frac{e^x}{\sqrt{1−e^{2t}}}\,dt$$

$$\displaystyle ∫\frac{e^x}{\sqrt{1−e^{2t}}}\,dt \quad = \quad \arcsin (e^t)+C$$

34) $$\displaystyle ∫\frac{e^t}{1+e^{2t}}\,dt$$

$$\displaystyle ∫\frac{e^t}{1+e^{2t}}\,dt = ∫\frac{e^t}{1+(e^t)^2}\,dt = ∫\frac{du}{1+u^2} \quad = \quad \arctan (e^t)+C$$

35) $$\displaystyle ∫\frac{dt}{t\sqrt{1−\ln^2t}}$$

$$\displaystyle ∫\frac{dt}{t\sqrt{1−\ln^2t}} \quad = \quad \arcsin (\ln t)+C$$

36) $$\displaystyle ∫\frac{dt}{t(1+\ln^2t)}$$

$$\displaystyle ∫\frac{dt}{t(1+\ln^2t)} \quad = \quad \arctan (\ln t)+C$$

37) $$\displaystyle ∫\frac{\arccos (2t)}{\sqrt{1−4t^2}}\,dt$$

$$\displaystyle ∫\frac{\arccos (2t)}{\sqrt{1−4t^2}}\,dt \quad = \quad −\frac{1}{2}(\arccos (2t))^2+C$$

38) $$\displaystyle ∫\frac{e^t\arccos (e^t)}{\sqrt{1−e^{2t}}}\,dt$$

$$\displaystyle ∫\frac{e^t\arccos (e^t)}{\sqrt{1−e^{2t}}}\,dt \quad = \quad \frac{\left(\arccos (e^t)\right)^2}{2}+C$$

In exercises 39 - 42, compute each definite integral.

39) $$\displaystyle ∫^{1/2}_0\frac{\tan(\arcsin t)}{\sqrt{1−t^2}}\,dt$$

$$\displaystyle ∫^{1/2}_0\frac{\tan(\arcsin t)}{\sqrt{1−t^2}}\,dt \quad = \quad \frac{1}{2}\ln\left(\frac{4}{3}\right)$$

40) $$\displaystyle ∫^{1/2}_{1/4}\frac{\tan(\arccos t)}{\sqrt{1−t^2}}\,dt$$

41) $$\displaystyle ∫^{1/2}_0\frac{\sin(\arctan t)}{1+t^2}\,dt$$

$$\displaystyle ∫^{1/2}_0\frac{\sin(\arctan t)}{1+t^2}\,dt \quad = \quad 1−\frac{2}{\sqrt{5}}$$

42) $$\displaystyle ∫^{1/2}_0\frac{\cos(\arctan t)}{1+t^2}\,dt$$

In exercises 43 - 50, compute each integral using appropriate substitutions and additional techniques.

43) $$\displaystyle ∫\frac{5}{x^2 + 10x + 34}\,dx$$

$$\displaystyle ∫\frac{5}{x^2 + 10x + 34}\,dx = 5∫\frac{1}{(x^2 + 10x + 25) + 9}\,dx = 5∫\frac{1}{(x+5)^2 + 3^2}\,dx \quad = \quad \frac{5}{3}\arctan\left( \frac{x+5}{3}\right) +C$$

44) $$\displaystyle ∫\frac{7}{x^2 - 2x + 5}\,dx$$

45) $$\displaystyle ∫\frac{2}{\sqrt{-x^2 + 8x + 3}}\,dx$$

$$\displaystyle ∫\frac{2}{\sqrt{-x^2 + 8x + 3}}\,dx = ∫\frac{2}{\sqrt{(\sqrt{19})^2-(x-4)^2}}\,dx \quad = \quad 2\arcsin \left( \frac{x-4}{\sqrt{19}} \right) + C$$

46) $$\displaystyle ∫\frac{dx}{\sqrt{-x^2 - 16x}}$$

47) $$\displaystyle ∫\frac{5x}{x^4 - 16x^2 + 100}\,dx$$

$$\displaystyle ∫\frac{5x}{x^4 - 16x^2 + 100}\,dx = \frac{5}{2}∫\frac{2x}{(x^2-8)^2 + 36}\,dx \quad = \quad \frac{5}{12}\arctan\left( \frac{x^2-8}{6} \right) + C$$

48) $$\displaystyle ∫\frac{x}{\sqrt{1-x^4}}\,dx$$

49) $$\displaystyle ∫\frac{8}{x \sqrt{x^4-9}}\,dx$$

$$\displaystyle ∫\frac{8}{x \sqrt{x^4-9}}\,dx = \frac{8}{2}∫\frac{2x}{x^2\sqrt{(x^2)^2 - 3^2}}\,dx \quad = \quad \frac{4}{3}\text{arcsec}\left( \frac{x^2}{3} \right) + C$$

50) $$\displaystyle ∫\frac{5}{x \sqrt{x^4-16x^2}}\,dx$$

$$\displaystyle ∫\frac{5}{x \sqrt{x^4-16x^2}}\,dx = ∫\frac{5}{|x|\sqrt{x^2 - 16}}\,dx \quad = \quad \frac{5}{4}\text{arcsec}\left( \frac{|x|}{4} \right) + C$$

51) $$\displaystyle ∫\frac{7x^3+5x}{x^4 +9}\,dx$$

$$\displaystyle ∫\frac{7x^3+5x}{x^4 +9}\,dx = \frac{7}{4}∫\frac{4x^3}{x^4 +9}\,dx+\frac{5}{2}∫\frac{2x}{(x^2)^2 +3^2}\,dx\quad = \quad \frac{7}{4}\ln|x^4 + 9| + \frac{5}{6}\arctan\left( \frac{x^2}{3} \right) + C$$

52) $$\displaystyle ∫\frac{2x+5}{x^2 +49}\,dx$$

53) $$\displaystyle ∫\frac{2x-1}{\sqrt{36-25x^2}}\,dx$$

$$\displaystyle ∫\frac{2x-1}{\sqrt{36-25x^2}}\,dx = \frac{2}{-50}∫\frac{-50x}{\sqrt{36-25x^2}}\,dx-\frac{1}{5}∫\frac{5}{\sqrt{6^2-(5x)^2}}\,dx\quad = \quad -\frac{2}{25}\sqrt{36-25x^2} - \frac{1}{5}\arcsin\left( \frac{5x}{6} \right) + C$$

54) $$\displaystyle\int \frac{3x^3+4x^2+2x-22}{x^2+3x+5}\,dx$$

55) $$\displaystyle\int \frac{-x^3+14x^2-46x-7}{x^2-7x+1}\,dx$$

Use long division first, since the degree of the numerator is not less than the degree of the denominator.
$$\displaystyle \int \frac{-x^3+14x^2-46x-7}{x^2-7x+1}\,dx = \int \left( -x + 7 + \frac{4x-14}{x^2-7x+1} \right) \,dx \quad = \quad -\frac{x^2}{2} + 7x +2\ln| x^2 - 7x + 1 | + C$$

56) $$\displaystyle\int \frac{x^3-x}{x^2+4x+9}\,dx$$

57) $$\displaystyle\int \frac{x^2+5x-2}{x^2-10x+32}\,dx$$

Use long division first, since the degree of the numerator is not less than the degree of the denominator.
$$\displaystyle \int \frac{x^2+5x-2}{x^2-10x+32}\,dx = \int \left( 1 + \frac{15x-34}{x^2-10x+32} \right) \,dx =\int \,dx+\frac{15}{2}\int \frac{2x-10}{x^2-10x+32} \,dx +41\int\frac{1}{(x-5)^2 + (\sqrt{7})^2}\,dx \quad = \quad x + \frac{15}{2} \ln| x^2-10x+32 | + \frac{41}{\sqrt{7}}\arctan\left( \frac{x-5}{\sqrt{7}} \right) + C$$

58) $$\displaystyle\int \frac{x^3}{x^2+9}\,dx$$

Use long division first, since the degree of the numerator is not less than the degree of the denominator. Fill in missing terms with zero-terms.
$$\displaystyle \int \frac{x^3}{x^2+9}\,dx \quad = \quad \frac{x^2}{2} -\frac{9}{2}\ln(x^2 +9) + C$$

59) $$\displaystyle\int \frac{\cos (x)}{\sin^2 (x)+1}\,dx$$

$$\displaystyle \int \frac{\cos (x)}{\sin^2 (x)+1}\,dx \quad = \quad \arctan(\sin x) + C$$

60) $$\displaystyle\int \frac{\cos (x)}{1-\sin^2 (x)}\,dx$$

\begin{align*} \displaystyle \int \frac{\cos (x)}{1-\sin^2 (x)}\,dx &= \int\frac{\cos x}{\cos^2 x}\,dx \\[5pt] &= \int \sec x\,dx = ∫\sec x\cdot\frac{\sec x + \tan x}{\sec x + \tan x}\,dx \\[5pt] &= ∫\frac{(\sec^2 x + \sec x\tan x)}{\tan x + \sec x}\,dx \\[5pt] &= \ln|\tan x + \sec x | + C \end{align*}

61) $$\displaystyle\int \frac{3x-3}{\sqrt{x^2-2x-6}}\,dx$$

$$\displaystyle \int \frac{3x-3}{\sqrt{x^2-2x-6}}\,dx \quad = \quad 3\sqrt{x^2-2x-6} + C$$

62) $$\displaystyle\int \frac{x-3}{\sqrt{x^2-6x+8}}\,dx$$

63) For $$A>0$$, compute $$\displaystyle I(A)=∫^{A}_{−A}\frac{dt}{1+t^2}$$ and evaluate $$\displaystyle \lim_{a→∞}I(A)$$, the area under the graph of $$\dfrac{1}{1+t^2}$$ on $$[−∞,∞]$$.

$$2\arctan (A)→π$$ as $$A→∞$$

64) For $$1<B<∞$$, compute $$\displaystyle I(B)=∫^B_1\frac{dt}{t\sqrt{t^2−1}}$$ and evaluate $$\displaystyle \lim_{B→∞}I(B)$$, the area under the graph of $$\frac{1}{t\sqrt{t^2−1}}$$ over $$[1,∞)$$.

65) Use the substitution $$u=\sqrt{2}\cot x$$ and the identity $$1+\cot^2x=\csc^2x$$ to evaluate $$\displaystyle ∫\frac{dx}{1+\cos^2x}$$. (Hint: Multiply the top and bottom of the integrand by $$\csc^2x$$.)

Using the hint, one has $$\displaystyle ∫\frac{\csc^2x}{\csc^2x+\cot^2x}\,dx=∫\frac{\csc^2x}{1+2\cot^2x}\,dx.$$ Set $$u=\sqrt{2}\cot x.$$ Then, $$du=−\sqrt{2}\csc^2x$$ and the integral is $$\displaystyle −\tfrac{1}{\sqrt{2}}∫\frac{du}{1+u^2}=−\tfrac{\sqrt{2}}{2}\arctan u+C=\tfrac{\sqrt{2}}{2}\arctan (\sqrt{2}\cot x)+C$$. If one uses the identity $$\arctan s+\arctan (\frac{1}{s})=\frac{π}{2}$$, then this can also be written $$\tfrac{\sqrt{2}}{2}\arctan (\frac{\tan x}{\sqrt{2}})+C.$$

66) [T] Approximate the points at which the graphs of $$f(x)=2x^2−1$$ and $$g(x)=(1+4x^2)^{−3/2}$$ intersect, and approximate the area between their graphs accurate to three decimal places.

67) [T] Approximate the points at which the graphs of $$f(x)=x^2−1$$ and $$f(x)=x^2−1$$ intersect, and approximate the area between their graphs accurate to three decimal places.

$$x≈±1.13525.$$ The left endpoint estimate with $$N=100$$ is 2.796 and these decimals persist for $$N=500$$.
68) Use the following graph to prove that $$\displaystyle ∫^x_0\sqrt{1−t^2}\,dt=\frac{1}{2}x\sqrt{1−x^2}+\frac{1}{2}\arcsin x.$$