Class XII Integration by Substitution (Ex 7.3, Part 3)

This lesson finishes off Exercise 7.3 with several more integrals, each solved by choosing a good substitution or by simplifying the integrand with a trig identity first. The problems are worked in the order they appear in the video.

An exponential integrand

Evaluate

$\int a^{x}\bigl(1 + x\ln a\bigr)\sec^{2}\!\bigl(a^{x}x\bigr)\,dx.$

Put

$t = a^{x}x.$

Differentiating, $dt = a^{x}\bigl(1 + x\ln a\bigr)\,dx$, which is exactly the factor sitting in front. So the integral becomes

$\int \sec^{2} t\,dt = \tan t + C = \tan\!\bigl(a^{x}x\bigr) + C.$

A fraction reduced to a rational function of tan

Evaluate $\displaystyle\int \frac{1}{\sin x \,\cos^{3} x}\,dx$. Multiply numerator and denominator by $\cos x$ and rewrite using $\tan x$:

$\int \frac{\cos x}{\sin x\,\cos^{4} x}\,dx = \int \frac{1}{\tan x}\,\sec^{2}x\;\sec^{2}x\,dx.$

Put $t = \tan x$, so $dt = \sec^{2}x\,dx$ and $\sec^{2}x = 1 + \tan^{2}x = 1 + t^{2}$:

$\int \frac{1 + t^{2}}{t}\,dt = \int \frac{1}{t}\,dt + \int t\,dt = \ln|t| + \tfrac{t^{2}}{2} + C.$

Back-substitute:

$\ln|\tan x| + \tfrac{\tan^{2} x}{2} + C.$

Integrating $\tan^{4} x$

Write $\tan^{4} x = \tan^{2}x\,\tan^{2}x$ and use $\tan^{2}x = \sec^{2}x - 1$:

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

First integral. Put $t = \tan x$, $dt = \sec^{2}x\,dx$:

$\int t^{2}\,dt = \tfrac{\tan^{3}x}{3}.$

Second integral. Again use $\tan^{2}x = \sec^{2}x - 1$:

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

Combining,

$\int \tan^{4}x\,dx = \tfrac{\tan^{3}x}{3} - \tan x + x + C.$

A fraction of cosines

Evaluate $\displaystyle\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha}\,dx$. Using $\cos 2\theta = 2\cos^{2}\theta - 1$,

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

The numerator is a difference of squares, $\cos^{2}x - \cos^{2}\alpha = (\cos x + \cos\alpha)(\cos x - \cos\alpha)$, so the $(\cos x - \cos\alpha)$ cancels:

$\int 2(\cos x + \cos\alpha)\,dx = 2\sin x + 2x\cos\alpha + C,$

remembering that $\cos\alpha$ is a constant.

$\dfrac{\sin^{2} x}{1 + \cos x}$

Use $\sin^{2}x = 1 - \cos^{2}x = (1 + \cos x)(1 - \cos x)$, so the $(1 + \cos x)$ cancels:

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

$\dfrac{\cos x}{1 - \cos x}$

Multiply top and bottom by the conjugate $1 + \cos x$:

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

Split into two pieces:

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

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

$-\csc x - \cot x - x + C.$

$\dfrac{1 - \cos x}{1 + \cos x}$

Use the half-angle forms $1 - \cos x = 2\sin^{2}\tfrac{x}{2}$ and $1 + \cos x = 2\cos^{2}\tfrac{x}{2}$:

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

Integrating,

$2\tan\tfrac{x}{2} - x + C.$

Product to sum: $\sin 3x\,\cos 4x$

Multiply and divide by $2$ and use $2\sin X\cos Y = \sin(X+Y) + \sin(X-Y)$:

$\int \sin 3x\,\cos 4x\,dx = \tfrac{1}{2}\int \bigl(\sin 7x + \sin(-x)\bigr)\,dx = \tfrac{1}{2}\int \bigl(\sin 7x - \sin x\bigr)\,dx.$

Integrate term by term:

$\tfrac{1}{2}\left(-\tfrac{\cos 7x}{7} + \cos x\right) + C = \frac{\cos x}{2} - \frac{\cos 7x}{14} + C.$

Key takeaways

• A good substitution often appears when the derivative of an inner expression is already a factor in the integrand.
• Identities such as $\tan^{2}x = \sec^{2}x - 1$ and $\cot^{2}x = \csc^{2}x - 1$ reduce higher powers to standard integrals.
• Multiplying by a conjugate, or spotting a difference of squares, lets a cancellation simplify the fraction before you integrate.
• Convert products of sines and cosines into sums with the product-to-sum formulas before integrating.