Integration by Substitution, Exercise 7.3 Part 1

This lesson works through the first questions of Exercise 7.3 from Class 12 Integration. The recurring ideas are to rewrite even powers of sine with the half-angle identity, and to integrate odd powers by splitting off one factor and substituting.

Example 1: $\int \sin^{2}(2x+5)\,dx$

Use the identity $\sin^{2}\theta = \dfrac{1-\cos 2\theta}{2}$ with $\theta = 2x+5$, so $2\theta = 4x+10$:

$\int \sin^{2}(2x+5)\,dx = \int \frac{1-\cos(4x+10)}{2}\,dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)\,dx.$

Since $\int \cos(4x+10)\,dx = \dfrac{\sin(4x+10)}{4}$,

$= \frac{x}{2} - \frac{1}{8}\sin(4x+10) + c.$

Example 2: $\int \sin^{3}(2x+1)\,dx$

Split off one factor of sine and write $\sin^{2}$ with the Pythagorean identity:

$\int \sin^{3}(2x+1)\,dx = \int \bigl(1-\cos^{2}(2x+1)\bigr)\sin(2x+1)\,dx.$

Put $t = \cos(2x+1)$, so $dt = -2\sin(2x+1)\,dx$, which gives $\sin(2x+1)\,dx = -\tfrac{1}{2}\,dt$. Then

$= \int (1-t^{2})\left(-\tfrac{1}{2}\right)dt = -\frac{1}{2}\left(t - \frac{t^{3}}{3}\right) + c.$

Replacing $t = \cos(2x+1)$,

$= -\frac{1}{2}\cos(2x+1) + \frac{1}{6}\cos^{3}(2x+1) + c.$

Example 3: $\int \sin^{3}x\,\cos^{3}x\,dx$

Keep one cosine for the substitution and turn the remaining $\cos^{2}x$ into $1-\sin^{2}x$:

$\int \sin^{3}x\,\cos^{3}x\,dx = \int \sin^{3}x\,(1-\sin^{2}x)\cos x\,dx.$

Put $t = \sin x$, so $dt = \cos x\,dx$:

$= \int t^{3}(1-t^{2})\,dt = \int \bigl(t^{3}-t^{5}\bigr)\,dt = \frac{t^{4}}{4} - \frac{t^{6}}{6} + c.$

Replacing $t = \sin x$,

$= \frac{\sin^{4}x}{4} - \frac{\sin^{6}x}{6} + c.$

Example 4: $\int \dfrac{\cos x}{1-\cos x}\,dx$

Multiply the top and bottom by $1+\cos x$ to use $1-\cos^{2}x = \sin^{2}x$:

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

Split the fraction:

$\int \frac{\cos x}{\sin^{2}x}\,dx + \int \frac{\cos^{2}x}{\sin^{2}x}\,dx = \int \cot x\,\csc x\,dx + \int \cot^{2}x\,dx.$

The first integral is $-\csc x$. For the second, use $\cot^{2}x = \csc^{2}x - 1$:

$\int \cot^{2}x\,dx = \int \bigl(\csc^{2}x - 1\bigr)\,dx = -\cot x - x.$

Adding the parts,

$= -\csc x - \cot x - x + c.$

Example 5: $\int \sin^{4}x\,dx$

Write $\sin^{4}x = \bigl(\sin^{2}x\bigr)^{2}$ and use the half-angle identity:

$\int \sin^{4}x\,dx = \int \left(\frac{1-\cos 2x}{2}\right)^{2}dx = \frac{1}{4}\int \bigl(1 - 2\cos 2x + \cos^{2}2x\bigr)\,dx.$

For the last term use $\cos^{2}2x = \dfrac{1+\cos 4x}{2}$. Integrating term by term,

$= \frac{1}{4}\left[\,x - \sin 2x + \frac{1}{2}\left(x + \frac{\sin 4x}{4}\right)\right] = \frac{1}{4}x - \frac{1}{4}\sin 2x + \frac{1}{8}x + \frac{1}{32}\sin 4x.$

Collecting the $x$ terms, $\tfrac{1}{4}x + \tfrac{1}{8}x = \tfrac{3}{8}x$, so

$= \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + c.$

Key takeaways

• For an even power of sine, rewrite it with the half-angle identity $\sin^{2}\theta = \tfrac{1-\cos 2\theta}{2}$ and integrate term by term.
• For an odd power of sine, peel off one sine factor and substitute $t = \cos x$; for an odd power of cosine, peel off one cosine factor and substitute $t = \sin x$.
• A fraction like $\tfrac{\cos x}{1-\cos x}$ becomes integrable after multiplying top and bottom by $1+\cos x$ to bring in $\sin^{2}x$.