JC H2 Mathematics: Integration Techniques

The same inside function gives three different integrals

Find the following indefinite integrals. In each, the inside function is $u=1+x^2$, with $du=2x\,dx$. The outside function determines whether the answer is a power, exponential or logarithm.

For $I=\int6x(1+x^2)^4\,dx$,

The logarithm needs no modulus here because $1+x^2>0$. In general, $\int f'/f=\ln|f|+C$ on an interval where $f\ne0$. The power formula excludes exponent −1.

Make the fraction proper, then use partial fractions

Find $\displaystyle\int\frac{x^2+1}{x^2-3x+2}\,dx$. The numerator and denominator have the same degree, so divide first.

Setting $x=1$ gives $A=-2$; setting $x=2$ gives $B=5$. Thus

This is valid on any interval avoiding $x=1,2$. Integrating across either pole as though the function were continuous is invalid.

Long division followed by completing the square

Find $\displaystyle\int\frac{x^2-x}{x^2-2x+3}\,dx$. Write $Q=x^2-2x+3=(x-1)^2+2$, so $Q'=2x-2$.

The derivative part produces a logarithm; the constant remainder produces inverse tangent.

Because $Q>0$, the logarithm is defined for every real $x$. If the numerator were constant, only the inverse-tangent integral would be needed.

A square-root quadratic denominator

Find $\displaystyle\int\frac{x+2}{\sqrt{8-2x-x^2}}\,dx$. Complete the square and split the numerator:

Put $u=x+1$. The two resulting standard forms are different:

The integrand requires $-4<x<2$. Differentiating both terms recovers the two numerator parts.

Odd powers and even powers use different identities

Find $\int\sin^3x\,dx$ and $\int(\sin^23x+\tan^23x)\,dx$.

For an odd sine power, keep one sine factor and convert the rest to cosine:

For the even power, use the double-angle formula. For tangent squared, use secant squared minus 1:

Retain the factors 1/6 and 1/3 introduced when integrating multiples of x. The answer is valid on intervals avoiding the tangent poles.

A trigonometric substitution with exact limits

Use $x=\sin^2\theta$, $0\leq\theta\leq\pi/2$, to evaluate $\displaystyle I=\int_0^{1/2}\sqrt{\frac{16x}{1-x}}\,dx$.

Sine and cosine are non-negative on the stated interval; that determines the square-root signs.

No substitution back to x is needed because the new limits have already been used.

A substitution that removes two different roots

For $x>0$, use $x=u^6$, $u>0$, to integrate $\displaystyle\frac1{\sqrt{x}+\sqrt[3]{x}}$.

Long division is still needed after the substitution:

For an indefinite integral, replace the variable back at the end. The denominator is non-zero on x > 0.

Differentiate inverse tangent; integrate the polynomial

Find $\displaystyle\int x\tan^{-1}(2x)\,dx$. Choose $u=\tan^{-1}(2x)$, $dv=x\,dx$.

The choice of u makes the remaining integral rational and simpler. For ln x or inverse sine alone, take dv = dx.

Integration by parts twice returns the original integral

Find $I=\int\cos(3\ln x)\,dx$, $x>0$. Put $u=\ln x$, so $dx=e^u\,du$.

Substitute the expression for J and collect I on one side:

Do not stop after the second integration by parts: the original integral has to be isolated.

Differentiate, rearrange, hence integrate

Let $F(x)=e^{\sin^{-1}x}\sqrt{1-x^2}$, $-1<x<1$. Differentiate F, then find $\displaystyle I=\int\frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx$.

Rearrange the derivative identity to match the requested integrand:

To find the remaining integral $J$, integrate by parts with $u=e^{\sin^{-1}x}$, $dv=dx$:

The “hence” instruction points to the derivative identity. It also helps organise the two linked integrals.

Use one antiderivative on both sides of the split

Find $\displaystyle\int_0^{\pi/2}|2x-1|\cos x\,dx$. First integrate $(2x-1)\cos x$ by parts.

The factor $2x-1$ changes sign at $x=1/2$. Cosine is non-negative throughout this interval.

The value is approximately 0.6313, and is positive. Taking the modulus of one final signed integral would give a different result.