Reducing an Integral to Standard Form by Simplification

Many trigonometric or surd integrands look intractable at first glance, yet collapse to an elementary formula once a suitable identity or algebraic trick is used. This section works through eleven such integrals, all of which support Exercise 3.1.

$\int\cos^3x\,dx$. Use the triple-angle identity $\cos3A=4\cos^3A-3\cos A$, rearranged as $\cos^3A=\frac14(\cos3A+3\cos A)$. So $I=\frac14\int(\cos3x+3\cos x)dx=\frac{1}{12}\sin3x+\frac34\sin x+c$.

$\int\sqrt{1+\sin3x}\,dx$. Write $1+\sin3x=\cos^2\frac{3x}2+\sin^2\frac{3x}2+2\sin\frac{3x}2\cos\frac{3x}2=\left(\cos\frac{3x}2+\sin\frac{3x}2\right)^2$, so the square root is $\left|\cos\frac{3x}2+\sin\frac{3x}2\right|$; taking the positive branch, $I=\frac23\sin\frac{3x}2-\cos\frac{3x}2+c$.

$\int\sin^4x\,dx$. Write $\sin^4x=(\sin^2x)^2=\left(\frac{1-\cos2x}2\right)^2=\frac14(1-2\cos2x+\cos^22x)$, then use $\cos^22x=\frac{1+\cos4x}2$ again to reduce everything to a linear combination of $1,\cos2x,\cos4x$: $I=\frac14\left(\frac32x-\sin2x+\frac18\sin4x\right)+c$.

$\int\sin5x\cos7x\,dx$. Use $2\sin A\cos B=\sin(A+B)+\sin(A-B)$: $I=\frac12\int[\sin12x-\sin2x]dx=-\frac{1}{24}\cos12x+\frac14\cos2x+c$.

$\int\dfrac{\sin^3x-\cos^3x}{\sin^2x\cos^2x}\,dx$. Split the fraction into $\frac{\sin x}{\cos^2x}-\frac{\cos x}{\sin^2x}=\sec x\tan x-\csc x\cot x$, giving $I=\sec x+\csc x+c$.

$\int\dfrac{1}{1-\sin x}\,dx$. Multiply top and bottom by $1+\sin x$: the denominator becomes $1-\sin^2x=\cos^2x$, so $I=\int\frac{1+\sin x}{\cos^2x}dx=\int(\sec^2x+\sec x\tan x)dx=\tan x+\sec x+c$.

$\int\dfrac{\cos x}{1-\cos x}\,dx$. Multiply by $\frac{1+\cos x}{1+\cos x}$: denominator becomes $\sin^2x$, numerator becomes $\cos x+\cos^2x$; splitting gives $\csc x\cot x+\cot^2x=\csc x\cot x+\csc^2x-1$, so $I=-\csc x-\cot x-x+c$.

$\int\dfrac{\cos x-\cos2x}{1-\cos x}\,dx$. Since $\cos2x=2\cos^2x-1$, the numerator is $\cos x-2\cos^2x+1=\cos x(1-\cos x)+(1-\cos x)(1+\cos x)$... more directly, $\cos x-\cos2x=\cos x+1-2\cos^2x=(1-\cos x)(1+2\cos x)$, so the fraction simplifies to $1+2\cos x$, giving $I=x+2\sin x+c$. …