P3 5.1 - Integrating standard functions | Edexcel A-Level International Mathematics

Extend integration beyond powers of $x$ to exponentials, trigonometric functions and reciprocals. Reversing a derivative explains both the shape of the answer and the scale factor in front of it.

Exponential integrals

Which function has derivative $e^{3x}$ ? The answer is close to $e^{3x}$ , but differentiating that expression produces three times too much. Integration must undo that factor.

You already know that an antiderivative differentiates back to the integrand, and that an indefinite integral includes an arbitrary constant. Since

\frac{d}{dx}e^{kx}=ke^{kx},

dividing by $k$ gives the required derivative. This works for a negative $k$ too.

Exponential integral

For a nonzero constant $k$ ,

\int e^{kx}\,dx=\frac1k e^{kx}+c.

For $k=0$ , the integrand is $1$ , so its integral is $x+c$ .

Not in the formula booklet: learn it.

A different positive base is another form of the same exponential. Since $a^x=e^{x\ln a}$ , with $a>0$ , its inner coefficient is $\ln a$ . Thus reversing $\frac{d}{dx}a^x=a^x\ln a$ gives the general-base rule.

Positive-base exponential integral

For $a>0$ and $a\ne1$ ,

\int a^x\,dx=\frac{a^x}{\ln a}+c.

For $a=1$ , the integrand is $1^x=1$ , so its integral is $x+c$ ; division by $\ln1=0$ is invalid.

Not in the formula booklet: learn it.

The sign of $\ln a$ matters: it is positive for $a>1$ and negative for $0<a<1$ . The following pair changes only the base to its reciprocal.

Worked example 1: a base greater than one

Find $\int4^x\,dx$ .

Plan. Divide by $\ln4$ , the constant multiplier in the derivative of $4^x$ .

\int4^x\,dx=\int e^{x\ln4}\,dx =\frac{e^{x\ln4}}{\ln4}+c.

Answer. $\frac{4^x}{\ln4}+c$ .

Check. Differentiating gives $\frac1{\ln4}\,4^x\ln4=4^x$ . At $x=0$ , both the required gradient and the checked derivative equal $1$ .

Change only the base from $4$ to $1/4$ . The logarithm in the denominator is now negative.

Worked example 2: a reciprocal base

Find $\int(1/4)^x\,dx$ .

Plan. Use the same rule, then simplify $\ln(1/4)=-\ln4$ .

\int(1/4)^x\,dx =\frac{(1/4)^x}{\ln(1/4)}+c =-\frac{(1/4)^x}{\ln4}+c.

Answer. $-\frac{(1/4)^x}{\ln4}+c$ .

Check. Its derivative is $-\frac1{\ln4}(1/4)^x\ln(1/4)=(1/4)^x$ . Despite the negative coefficient in the antiderivative, its derivative is positive, as the integrand requires.

Return to base $e$ , where $\ln e=1$ . A constant multiplier stays outside the integration. Choose the factor that cancels the multiplier produced by differentiation.

Worked example 3: positive exponent coefficient

Find $\int 3e^{2x}\,dx$ .

Plan. Keep the exponential and divide its coefficient by $2$ .

\int 3e^{2x}\,dx=3\left(\frac12e^{2x}\right)+c.

Why divide? Differentiating $e^{2x}$ multiplies by $2$ , so integration cancels that multiplication.

Answer. $\frac32e^{2x}+c$ .

Check. $\frac{d}{dx}(\frac32e^{2x}+c)=\frac32(2e^{2x})=3e^{2x}$ .

Change only the sign of the exponent coefficient. The integral's coefficient must change sign as well.

Worked example 4: negative exponent coefficient

Find $\int 3e^{-2x}\,dx$ .

Plan. Divide by the inner derivative $-2$ .

\int 3e^{-2x}\,dx=\frac{3}{-2}e^{-2x}+c.

The negative coefficient compensates for the negative derivative of $-2x$ .

Answer. $-\frac32e^{-2x}+c$ .

Check. Differentiation gives $(-\frac32)(-2)e^{-2x}=3e^{-2x}$ .

The exponent's form survives integration; only the multiplier changes.

Watch out

$\int e^{3x}\,dx$ is $\frac13e^{3x}+c$ , not $3e^{3x}+c$ . Differentiation multiplies by the inner derivative; integration here divides by it.

Use the same cancellation in this integral.

Sine and cosine integrals

Sine and cosine exchange roles under differentiation, with one sign change. Their integrals reverse those derivatives, with angles in radians.

Since the derivative of $\cos kx$ is $-k\sin kx$ , a sine integral needs a negative cosine. The derivative of $\sin kx$ is $k\cos kx$ , so a cosine integral needs a positive sine.

Sine and cosine integrals

For a nonzero constant $k$ , with angles in radians,

\int\sin kx\,dx=-\frac1k\cos kx+c, \qquad \int\cos kx\,dx=\frac1k\sin kx+c.

Not in the formula booklet: learn them.

Decide the function and sign before calculating its coefficient.

Worked example 5: integrating sine

Find $\int 6\sin3x\,dx$ .

Plan. Use a negative cosine, then divide the coefficient by $3$ .

\int6\sin3x\,dx=-\frac63\cos3x+c.

The minus sign reverses the sign in the derivative of cosine.

Answer. $-2\cos3x+c$ .

Check. $\frac{d}{dx}(-2\cos3x+c)=(-2)(-3\sin3x)=6\sin3x$ .

Change only sine to cosine. The scale factor is unchanged, but the new antiderivative is a positive sine.

Worked example 6: integrating cosine

Find $\int 6\cos3x\,dx$ .

Plan. Use a sine and divide by its inner derivative $3$ .

\int6\cos3x\,dx=\frac63\sin3x+c.

No negative sign is needed because the derivative of sine is positive cosine.

Answer. $2\sin3x+c$ .

Check. $\frac{d}{dx}(2\sin3x+c)=2(3\cos3x)=6\cos3x$ .

These rules require an angle that is a constant multiple of $x$ . A squared sine is a different integrand.

Watch out

$\sin(3x)$ means sine of $3x$ ; $\sin^2x$ means $(\sin x)^2$ . Do not apply the sine rule directly to a squared sine. You will learn to rewrite such integrands using identities.

Keep your calculator in radians whenever you use numerical values to check these derivatives or integrals.

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P3 5.1 - Integrating standard functions

Exponential integrals

Sine and cosine integrals

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