Indefinite Integrals and Substitution — Worked Solutions

Ques­tion 1: Inte­grat­ing with the stan­dard table#

The prob­lem#

Find: (a) $\displaystyle \int(5x^4 + 2x - 7)\,dx$; (b) $\displaystyle \int\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2dx$; (c) $\displaystyle \int(2\sin x - 3\sec^2 x)\,dx$.

Under­stand­ing the prob­lem#

Each inte­grand is a sum of sim­ple terms (in (b), after expand­ing). You need a func­tion whose deriv­a­tive is the inte­grand, plus the con­stant $C$.

The idea#

Use lin­ear­ity, $\displaystyle \int (af + bg)\,dx = a\int f\,dx + b\int g\,dx$, and inte­grate term by term from the table of stan­dard inte­grals. In (b), first expand the square so every term is a power of $x$.

Step-by-step solu­tion#

Part (a)

Step 1. Inte­grate each term with $\displaystyle \int x^n\,dx = \tfrac{x^{n+1}}{n + 1}$.

$$\displaystyle \int 5x^4\,dx = 5\cdot\frac{x^5}{5} = x^5, \quad \int 2x\,dx = 2\cdot\frac{x^2}{2} = x^2, \quad \int 7\,dx = 7x$$

Step 2. Com­bine and add one con­stant.

$$\displaystyle \int(5x^4 + 2x - 7)\,dx = x^5 + x^2 - 7x + C$$

Part (b)

Step 1. Expand the square using $(a + b)^2 = a^2 + 2ab + b^2$. Note $\displaystyle \sqrt{x} \cdot \tfrac{1}{\sqrt{x}} = 1$.

$$\displaystyle \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + 2 + \frac{1}{x}$$

Step 2. Inte­grate each term. The last term uses $\displaystyle \int \tfrac{1}{x}\,dx = \log\lvert x \rvert$.

$$\displaystyle \int\left(x + 2 + \frac{1}{x}\right)dx = \frac{x^2}{2} + 2x + \log\lvert x \rvert + C$$

Part (c)

Step 1. Use $\displaystyle \int \sin x\,dx = -\cos x$ and $\displaystyle \int \sec^2 x\,dx = \tan x$.

$$\displaystyle \int(2\sin x - 3\sec^2 x)\,dx = 2(-\cos x) - 3\tan x + C = -2\cos x - 3\tan x + C$$

Check­ing the answer#

Dif­fer­en­ti­ate each result. (a) $5x^4 + 2x - 7$. (b) $\displaystyle x + 2 + \tfrac{1}{x}$, which is the expanded inte­grand. (c) $2\sin x - 3\sec^2 x$. All match.

Answer#

(a) $x^5 + x^2 - 7x + C$; (b) $\displaystyle \frac{x^2}{2} + 2x + \log\lvert x \rvert + C$; (c) $-2\cos x - 3\tan x + C$.

Com­mon mis­take to avoid#

In (b), $\displaystyle \int x^{-1}\,dx$ is not $\displaystyle \tfrac{x^0}{0}$; the power rule fails for $n = -1$, and the answer is $\log\lvert x \rvert$.

Ques­tion 2: Find­ing a func­tion from its deriv­a­tive#

The prob­lem#

Find $f$ if $f'(x) = 3\cos x - 2e^x$ and $f(0) = 4$.

Under­stand­ing the prob­lem#

You know the deriv­a­tive of $f$ and one value of $f$. Inte­grat­ing gives $f$ up to a con­stant; the value $f(0) = 4$ fixes that con­stant. This is like Exam­ple 3.

The idea#

Inte­grate $f'(x)$ to get $f(x) = (\text{antiderivative}) + C$, then sub­sti­tute $x = 0$ to find $C$.

Step-by-step solu­tion#

Step 1. Inte­grate term by term, using $\displaystyle \int\cos x\,dx = \sin x$ and $\displaystyle \int e^x\,dx = e^x$.

$$\displaystyle f(x) = \int(3\cos x - 2e^x)\,dx = 3\sin x - 2e^x + C$$

Step 2. Use $f(0) = 4$. Remem­ber $\sin 0 = 0$ and $e^0 = 1$.

$$f(0) = 3(0) - 2(1) + C = -2 + C = 4$$

Step 3. Solve for $C$: $C = 6$.

Step 4. Write the final func­tion.

$$f(x) = 3\sin x - 2e^x + 6$$

Check­ing the answer#

$f'(x) = 3\cos x - 2e^x$ ✓, and $f(0) = 0 - 2 + 6 = 4$ ✓.

Answer#

$f(x) = 3\sin x - 2e^x + 6$.

Ques­tion 3: Sub­sti­tu­tion when a func­tion and its deriv­a­tive both appear#

The prob­lem#

Find: (a) $\displaystyle \int\frac{\sin(\log x)}{x}\,dx$; (b) $\displaystyle \int\frac{\cos x}{\sqrt{\sin x}}\,dx$; (c) $\displaystyle \int xe^{x^2}\,dx$.

Under­stand­ing the prob­lem#

None of these is in the stan­dard table as it stands. In each, look for an "inner" func­tion whose deriv­a­tive also appears as a fac­tor.

The idea#

The lesson's rule: if the inte­grand con­tains a func­tion and its deriv­a­tive, put $t$ equal to that func­tion. Then $dt$ absorbs the deriv­a­tive, and what is left is a stan­dard inte­gral in $t$.

• (a) $\log x$ is inside, and its deriv­a­tive $\displaystyle \tfrac{1}{x}$ is present.
• (b) $\sin x$ is inside, and its deriv­a­tive $\cos x$ is present.
• (c) $x^2$ is inside, and its deriv­a­tive $2x$ is present apart from the con­stant $2$.

Step-by-step solu­tion#

Part (a)

Step 1. Put $t = \log x$. Then $\displaystyle dt = \tfrac{1}{x}\,dx$.

Step 2. Rewrite: $\displaystyle \tfrac{\sin(\log x)}{x}\,dx = \sin t\,dt$.

$$\displaystyle \int \sin t\,dt = -\cos t + C$$

Step 3. Sub­sti­tute back $t = \log x$.

$$\displaystyle \int\frac{\sin(\log x)}{x}\,dx = -\cos(\log x) + C$$

Part (b)

Step 1. Put $t = \sin x$. Then $dt = \cos x\,dx$.

Step 2. Rewrite and inte­grate as a power, $t^{-1/2}$.

$$\displaystyle \int \frac{dt}{\sqrt{t}} = \int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$$

Step 3. Sub­sti­tute back.

$$\displaystyle \int\frac{\cos x}{\sqrt{\sin x}}\,dx = 2\sqrt{\sin x} + C$$

Part (c)

Step 1. Put $t = x^2$. Then $dt = 2x\,dx$, so $\displaystyle x\,dx = \tfrac{1}{2}dt$.

Step 2. Rewrite and inte­grate.

$$\displaystyle \int e^{t}\cdot\frac{1}{2}\,dt = \frac{1}{2}e^{t} + C$$

Step 3. Sub­sti­tute back.

$$\displaystyle \int xe^{x^2}\,dx = \frac{e^{x^2}}{2} + C$$

Check­ing the answer#

Dif­fer­en­ti­ate with the chain rule. (a) $\displaystyle \sin(\log x)\cdot\tfrac{1}{x}$ ✓. (b) $\displaystyle 2\cdot\tfrac{1}{2\sqrt{\sin x}}\cdot\cos x = \tfrac{\cos x}{\sqrt{\sin x}}$ ✓. (c) $\displaystyle \tfrac{1}{2}e^{x^2}\cdot 2x = xe^{x^2}$ ✓.

Answer#

(a) $-\cos(\log x) + C$; (b) $2\sqrt{\sin x} + C$; (c) $\displaystyle \frac{e^{x^2}}{2} + C$.

Com­mon mis­take to avoid#

For­get­ting to sub­sti­tute back. The final answer must be in terms of $x$, not $t$.

Ques­tion 4: More sub­sti­tu­tions, includ­ing the log form#

The prob­lem#

Find: (a) $\displaystyle \int\frac{2x + 3}{x^2 + 3x + 7}\,dx$; (b) $\displaystyle \int\tan^3 x\sec^2 x\,dx$; (c) $\displaystyle \int\frac{1}{x\log x}\,dx$.

Under­stand­ing the prob­lem#

Again look for an inner func­tion and its deriv­a­tive. In (a) and (c) the deriv­a­tive of the denom­i­na­tor sits in the numer­a­tor; in (b), $\sec^2 x$ is the deriv­a­tive of $\tan x$.

The idea#

Sub­sti­tute $t$ = the inner func­tion. A very use­ful spe­cial case (Exam­ple 6) is

$$\displaystyle \int\frac{f'(x)}{f(x)}\,dx = \log\lvert f(x) \rvert + C,$$

because with $t = f(x)$ the inte­gral becomes $\displaystyle \int\tfrac{dt}{t}$.

Step-by-step solu­tion#

Part (a)

Step 1. Notice $\displaystyle \tfrac{d}{dx}(x^2 + 3x + 7) = 2x + 3$, exactly the numer­a­tor. Put $t = x^2 + 3x + 7$, $dt = (2x + 3)\,dx$.

Step 2. Inte­grate.

$$\displaystyle \int\frac{dt}{t} = \log\lvert t \rvert + C = \log\lvert x^2 + 3x + 7 \rvert + C$$

Part (b)

Step 1. Put $t = \tan x$, so $dt = \sec^2 x\,dx$.

Step 2. The inte­gral becomes a power of $t$.

$$\displaystyle \int t^3\,dt = \frac{t^4}{4} + C = \frac{\tan^4 x}{4} + C$$

Part (c)

Step 1. Write the inte­grand as $\displaystyle \frac{1/x}{\log x}$. The numer­a­tor $\displaystyle \tfrac{1}{x}$ is the deriv­a­tive of $\log x$. Put $t = \log x$, $\displaystyle dt = \tfrac{1}{x}\,dx$.

Step 2. Inte­grate.

$$\displaystyle \int\frac{dt}{t} = \log\lvert t \rvert + C = \log\lvert \log x \rvert + C$$

Check­ing the answer#

(a) $\displaystyle \tfrac{2x + 3}{x^2 + 3x + 7}$ ✓. (b) $\displaystyle \tfrac{4\tan^3 x\sec^2 x}{4}$ ✓. (c) $\displaystyle \tfrac{1}{\log x}\cdot\tfrac{1}{x}$ ✓.

Answer#

(a) $\log\lvert x^2 + 3x + 7 \rvert + C$; (b) $\displaystyle \frac{\tan^4 x}{4} + C$; (c) $\log\lvert\log x\rvert + C$.

Ques­tion 5: Using trigono­met­ric iden­ti­ties#

The prob­lem#

Find: (a) $\displaystyle \int\cos^2 3x\,dx$; (b) $\displaystyle \int\sin 4x\sin 2x\,dx$; (c) $\displaystyle \int\cos^3 x\,dx$.

Under­stand­ing the prob­lem#

Pow­ers and prod­ucts of trigono­met­ric func­tions are not in the table. You must first rewrite each inte­grand as a sum of sim­ple sines and cosines, or as some­thing ready for sub­sti­tu­tion.

The idea#

• (a) Lower the power with $\displaystyle \cos^2\theta = \tfrac{1 + \cos 2\theta}{2}$ (as in Exam­ple 8).
• (b) Turn the prod­uct into a sum with $2\sin A\sin B = \cos(A - B) - \cos(A + B)$ (as in Exam­ple 9).
• (c) For an odd power, split off one fac­tor and use $\cos^2 x = 1 - \sin^2 x$, then sub­sti­tute $t = \sin x$ (as in Exam­ple 10).

Step-by-step solu­tion#

Part (a)

Step 1. Apply the iden­tity with $\theta = 3x$, so $2\theta = 6x$.

$$\displaystyle \cos^2 3x = \frac{1 + \cos 6x}{2}$$

Step 2. Inte­grate. Use $\displaystyle \int\cos kx\,dx = \tfrac{\sin kx}{k}$.

$$\displaystyle \int\frac{1 + \cos 6x}{2}\,dx = \frac{x}{2} + \frac{1}{2}\cdot\frac{\sin 6x}{6} + C = \frac{x}{2} + \frac{\sin 6x}{12} + C$$

Part (b)

Step 1. Use $\displaystyle \sin A\sin B = \tfrac{1}{2}\left[\cos(A - B) - \cos(A + B)\right]$ with $A = 4x$, $B = 2x$.

$$\displaystyle \sin 4x\sin 2x = \frac{1}{2}\left(\cos 2x - \cos 6x\right)$$

Step 2. Inte­grate each cosine.

$$\displaystyle \frac{1}{2}\left(\frac{\sin 2x}{2} - \frac{\sin 6x}{6}\right) + C = \frac{\sin 2x}{4} - \frac{\sin 6x}{12} + C$$

Part (c)

Step 1. Split off one $\cos x$ and rewrite the rest.

$$\cos^3 x = \cos^2 x\cdot\cos x = (1 - \sin^2 x)\cos x$$

Step 2. Put $t = \sin x$, $dt = \cos x\,dx$.

$$\displaystyle \int(1 - t^2)\,dt = t - \frac{t^3}{3} + C$$

Step 3. Sub­sti­tute back.

$$\displaystyle \int\cos^3 x\,dx = \sin x - \frac{\sin^3 x}{3} + C$$

Check­ing the answer#

(a) Deriv­a­tive: $\displaystyle \tfrac{1}{2} + \tfrac{\cos 6x}{2} = \cos^2 3x$ ✓. (b) Deriv­a­tive: $\displaystyle \tfrac{\cos 2x - \cos 6x}{2} = \sin 4x\sin 2x$ ✓. (c) Deriv­a­tive: $\cos x - \sin^2 x\cos x = \cos x(1 - \sin^2 x) = \cos^3 x$ ✓.

Answer#

(a) $\displaystyle \frac{x}{2} + \frac{\sin 6x}{12} + C$; (b) $\displaystyle \frac{\sin 2x}{4} - \frac{\sin 6x}{12} + C$; (c) $\displaystyle \sin x - \frac{\sin^3 x}{3} + C$.

Com­mon mis­take to avoid#

In (a), $\displaystyle \int\cos 6x\,dx = \tfrac{\sin 6x}{6}$, not $\sin 6x$. Divid­ing by the coef­fi­cient of $x$ is easy to for­get.

Ques­tion 6: Ratio­nal­is­ing a trigono­met­ric denom­i­na­tor#

The prob­lem#

Find $\displaystyle \int\frac{dx}{1 + \cos x}$ (mul­ti­ply by $1 - \cos x$).

Under­stand­ing the prob­lem#

The denom­i­na­tor $1 + \cos x$ is not a stan­dard form. The hint tells you to mul­ti­ply the numer­a­tor and denom­i­na­tor by $1 - \cos x$, which is like mul­ti­ply­ing by a con­ju­gate.

The idea#

$(1 + \cos x)(1 - \cos x) = 1 - \cos^2 x = \sin^2 x$. After that, the frac­tion splits into two stan­dard inte­grals: $\csc^2 x$ and $\csc x\cot x$.

Step-by-step solu­tion#

Step 1. Mul­ti­ply top and bot­tom by $1 - \cos x$.

$$\displaystyle \frac{1}{1 + \cos x} = \frac{1 - \cos x}{(1 + \cos x)(1 - \cos x)} = \frac{1 - \cos x}{1 - \cos^2 x} = \frac{1 - \cos x}{\sin^2 x}$$

Step 2. Split into two frac­tions.

$$\displaystyle \frac{1 - \cos x}{\sin^2 x} = \frac{1}{\sin^2 x} - \frac{\cos x}{\sin^2 x} = \csc^2 x - \csc x\cot x$$

(the sec­ond because $\displaystyle \tfrac{\cos x}{\sin^2 x} = \tfrac{1}{\sin x}\cdot\tfrac{\cos x}{\sin x}$).

Step 3. Inte­grate using $\displaystyle \int\csc^2 x\,dx = -\cot x$ and $\displaystyle \int\csc x\cot x\,dx = -\csc x$.

$$\displaystyle \int(\csc^2 x - \csc x\cot x)\,dx = -\cot x - (-\csc x) + C = -\cot x + \csc x + C$$

Check­ing the answer#

Dif­fer­en­ti­ate: $\displaystyle \csc^2 x - \csc x\cot x = \tfrac{1 - \cos x}{\sin^2 x} = \tfrac{1}{1 + \cos x}$ ✓. (Using half-angle for­mu­las, $\displaystyle \csc x - \cot x = \tan\tfrac{x}{2}$, another cor­rect form of the same answer.)

Answer#

$\displaystyle \int\frac{dx}{1 + \cos x} = \csc x - \cot x + C$.

Ques­tion 7: Sub­sti­tut­ing an inverse trigono­met­ric func­tion#

The prob­lem#

Find $\displaystyle \int\frac{\sin^{-1}x}{\sqrt{1 - x^2}}\,dx$.

Under­stand­ing the prob­lem#

The inte­grand is $\sin^{-1}x$ mul­ti­plied by $\displaystyle \tfrac{1}{\sqrt{1 - x^2}}$. From the stan­dard table, $\displaystyle \tfrac{1}{\sqrt{1 - x^2}}$ is the deriv­a­tive of $\sin^{-1}x$.

The idea#

A func­tion ($\sin^{-1}x$) and its deriv­a­tive are both present, so sub­sti­tute $t = \sin^{-1}x$, just as in Exam­ple 5.

Step-by-step solu­tion#

Step 1. Put $t = \sin^{-1}x$. Then $\displaystyle dt = \frac{1}{\sqrt{1 - x^2}}\,dx$.

Step 2. The inte­gral becomes

$$\displaystyle \int t\,dt = \frac{t^2}{2} + C$$

Step 3. Sub­sti­tute back.

$$\displaystyle \int\frac{\sin^{-1}x}{\sqrt{1 - x^2}}\,dx = \frac{(\sin^{-1}x)^2}{2} + C$$

Check­ing the answer#

Dif­fer­en­ti­ate: $\displaystyle \tfrac{2\sin^{-1}x}{2}\cdot\tfrac{1}{\sqrt{1 - x^2}}$ ✓.

Answer#

$\displaystyle \frac{(\sin^{-1}x)^2}{2} + C$.

Ques­tion 8: Sub­sti­tu­tion with a con­stant fac­tor to adjust#

The prob­lem#

Find $\displaystyle \int\sqrt{\sin 2x}\,\cos 2x\,dx$.

Under­stand­ing the prob­lem#

$\sin 2x$ sits inside the square root, and $\cos 2x$ is present. The deriv­a­tive of $\sin 2x$ is $2\cos 2x$, so the deriv­a­tive is present apart from a fac­tor of $2$.

The idea#

Put $t = \sin 2x$ and adjust for the miss­ing con­stant $2$. Then inte­grate a power of $t$.

Step-by-step solu­tion#

Step 1. Put $t = \sin 2x$. By the chain rule, $dt = 2\cos 2x\,dx$, so $\displaystyle \cos 2x\,dx = \tfrac{1}{2}dt$.

Step 2. Rewrite the inte­gral.

$$\displaystyle \int\sqrt{t}\cdot\frac{1}{2}\,dt = \frac{1}{2}\int t^{1/2}\,dt$$

Step 3. Inte­grate the power.

$$\displaystyle \frac{1}{2}\cdot\frac{t^{3/2}}{3/2} + C = \frac{1}{2}\cdot\frac{2}{3}t^{3/2} + C = \frac{t^{3/2}}{3} + C$$

Step 4. Sub­sti­tute back.

$$\displaystyle \int\sqrt{\sin 2x}\,\cos 2x\,dx = \frac{(\sin 2x)^{3/2}}{3} + C$$

Check­ing the answer#

Dif­fer­en­ti­ate: $\displaystyle \tfrac{1}{3}\cdot\tfrac{3}{2}(\sin 2x)^{1/2}\cdot 2\cos 2x = \sqrt{\sin 2x}\,\cos 2x$ ✓.

Answer#

$\displaystyle \frac{(\sin 2x)^{3/2}}{3} + C$.

Com­mon mis­take to avoid#

Writ­ing $dt = \cos 2x\,dx$ and los­ing the fac­tor $2$ from the chain rule; that gives an answer twice too big.