Integration by Substitution, Exercise 7.3 Part 2

This lesson continues Exercise 7.3 from Class 12 Integration. The recurring idea is to reshape each integrand with a trigonometric identity so it becomes a standard result or a simple substitution.

Example 1: $\int \dfrac{\sin^{2}x}{1-\cos x}\,dx$

Write $\sin^{2}x = 1-\cos^{2}x$ and factor it as a difference of squares:

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

Integrating term by term,

$= x + \sin x + c.$

If the denominator were $1+\cos x$ instead, the same factoring leaves $\int (1-\cos x)\,dx = x - \sin x + c$.

Example 2: $\int \dfrac{\cos x - \sin x}{1+\sin 2x}\,dx$

Use the identity $1+\sin 2x = (\cos x + \sin x)^{2}$:

$\int \frac{\cos x - \sin x}{(\cos x + \sin x)^{2}}\,dx.$

Put $t = \cos x + \sin x$, so $dt = (-\sin x + \cos x)\,dx = (\cos x - \sin x)\,dx$. The numerator is exactly $dt$:

$= \int \frac{dt}{t^{2}} = \int t^{-2}\,dt = -\frac{1}{t} + c = -\frac{1}{\cos x + \sin x} + c.$

Example 3: $\int \tan^{3}2x\,\sec 2x\,dx$

Keep one factor of $\tan 2x\,\sec 2x$ and turn the remaining $\tan^{2}2x$ into $\sec^{2}2x - 1$:

$\int \tan^{3}2x\,\sec 2x\,dx = \int \bigl(\sec^{2}2x - 1\bigr)\,\sec 2x\,\tan 2x\,dx.$

Put $t = \sec 2x$, so $dt = 2\sec 2x\,\tan 2x\,dx$, which gives $\sec 2x\,\tan 2x\,dx = \tfrac{1}{2}\,dt$:

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

Replacing $t = \sec 2x$,

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

Example 4: $\int \dfrac{\sin^{3}x + \cos^{3}x}{\sin^{2}x\,\cos^{2}x}\,dx$

Split the fraction into two and cancel:

$\int \frac{\sin^{3}x}{\sin^{2}x\,\cos^{2}x}\,dx + \int \frac{\cos^{3}x}{\sin^{2}x\,\cos^{2}x}\,dx = \int \frac{\sin x}{\cos^{2}x}\,dx + \int \frac{\cos x}{\sin^{2}x}\,dx.$

Rewrite each as a standard product:

$= \int \tan x\,\sec x\,dx + \int \cot x\,\csc x\,dx.$

Since $\int \tan x\,\sec x\,dx = \sec x$ and $\int \cot x\,\csc x\,dx = -\csc x$,

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

Example 5: $\int \dfrac{\cos 2x + 2\sin^{2}x}{\cos^{2}x}\,dx$

Write $\cos 2x = 1 - 2\sin^{2}x$ in the numerator:

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

So the integral collapses to

$\int \frac{1}{\cos^{2}x}\,dx = \int \sec^{2}x\,dx = \tan x + c.$

Example 6: $\int \dfrac{\cos 2x}{(\cos x + \sin x)^{2}}\,dx$

Use $\cos 2x = \cos^{2}x - \sin^{2}x = (\cos x + \sin x)(\cos x - \sin x)$ and cancel one factor:

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

Here the numerator is the derivative of the denominator, since $\dfrac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x$. Using $\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)|$,

$= \log|\cos x + \sin x| + c.$

Example 7: $\int \sin^{-1}(\cos x)\,dx$

Write $\cos x = \sin\!\left(\tfrac{\pi}{2} - x\right)$, so $\sin^{-1}(\cos x) = \sin^{-1}\sin\!\left(\tfrac{\pi}{2} - x\right) = \tfrac{\pi}{2} - x$:

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

Example 8: $\int \dfrac{\sin^{2}x - \cos^{2}x}{\sin^{2}x\,\cos^{2}x}\,dx$

Split the fraction and cancel:

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

These are standard:

$= \int \sec^{2}x\,dx - \int \csc^{2}x\,dx = \tan x - (-\cot x) = \tan x + \cot x + c.$

Key takeaways

• Factoring a difference of squares in the numerator often cancels the denominator and leaves a simple integral.
• The identity $1+\sin 2x = (\cos x + \sin x)^{2}$ turns the second example into the substitution $t = \cos x + \sin x$.
• Double-angle identities such as $\cos 2x = 1 - 2\sin^{2}x$ and $\cos 2x = \cos^{2}x - \sin^{2}x$ simplify the numerator before integrating.
• When the numerator is the derivative of the denominator, the integral is $\log$ of the denominator.