Lecture 28 Notes: Integration by Substitution with Inverse Trigonometric Functions✅

Integration by Substitution: Indefinite Integrals & Inverse Trigonometric Functions

If we have an integral of the form: I = \int f'(g(x))g'(x) dx, it equals f(g(x)) + c.
This is because differentiating f(g(x)) yields f'(g(x))g'(x).

Consider integrands involving products of powers of \cos and \sin, or \tan and \sec.
Define a new variable u as one of these trigonometric functions.
Utilize identities like \sin^2(x) + \cos^2(x) = 1 or \tan^2(x) + 1 = \sec^2(x).

Aim: Evaluate \int \sin^3(x) \cos^2(x) dx
Express \sin^3(x) as \sin(x) \sin^2(x)
Rewrite the integral as: \int \sin(x) \sin^2(x) \cos^2(x) dx
Note that \cos^2(x) means (\cos(x))^2 and \sin^2(x) means (\sin(x))^2
Let u = \cos(x). Then \frac{du}{dx} = -\sin(x), so du = -\sin(x) dx.
Express \sin^2(x) as 1 - \cos^2(x) using the trigonometric identity.
Rewrite the integral: \int \sin(x)(1 - \cos^2(x))\cos^2(x) dx
Substitute u = \cos(x). The integral becomes: \int (1 - u^2)u^2 (-du)
Simplify: -\int (u^2 - u^4) du
Integrate: -\left(\frac{u^3}{3} - \frac{u^5}{5}\right) + c
Substitute back u = \cos(x) to get: -\frac{\cos^3(x)}{3} + \frac{\cos^5(x)}{5} + c

Consider integrals with expressions like \sqrt{a^2 \pm x^2}, where a is a positive constant.
For \sqrt{a^2 - x^2}, substitute x = a \sin(u).
Then, \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2(u)} = \sqrt{a^2(1 - \sin^2(u))} = a\sqrt{\cos^2(u)} = a|\cos(u)|

Evaluate \int_{0}^{1} \frac{x^3}{\sqrt{1 - x^2}} dx
Let x = \sin(u). Then dx = \cos(u) du.
When x = 0, u = \arcsin(0) = 0. When x = 1, u = \arcsin(1) = \frac{\pi}{2}.
Substitute in the integral to get: \int{0}^{\frac{\pi}{2}} \frac{\sin^3(u)}{\sqrt{1 - \sin^2(u)}} \cos(u) du = \int{0}^{\frac{\pi}{2}} \sin^3(u) \cos(u) \frac{1}{|\cos(u)|} du
Since \cos(u) is positive on \left[0, \frac{\pi}{2}\right], we have: \int_{0}^{\frac{\pi}{2}} \sin^3(u) \cos(u)du
Use the antiderivative found above, F(u) = -\frac{\cos^3(u)}{3} + \frac{\cos^5(u)}{5}.
Evaluate at the limits: F\left(\frac{\pi}{2}\right) - F(0) = \left(-\frac{\cos^3(\frac{\pi}{2})}{3} + \frac{\cos^5(\frac{\pi}{2})}{5}\right) - \left(-\frac{\cos^3(0)}{3} + \frac{\cos^5(0)}{5}\right) = 0 - \left(-\frac{1}{3} + \frac{1}{5}\right) = \frac{2}{15}
Alternative: Substitute back u = \arcsin(x) and use original limits but it's more complex.

If the integrand contains a^2 + x^2 (without a square root), use the substitution x = a \tan(u).
Recall the identity: 1 + \tan^2(x) = \sec^2(x). Multiplying by a^2 gives: a^2 + a^2\tan^2(x) = a^2\sec^2(x).
Specifically, consider \int \frac{1}{a^2 + x^2} dx
Let x = a \tan(u). Then dx = a \sec^2(u) du.
Substitute to get: \int \frac{1}{a^2 + a^2\tan^2(u)} a \sec^2(u) du
Simplify: \int \frac{a \sec^2(u)}{a^2\sec^2(u)} du = \frac{1}{a} \int du = \frac{1}{a}{u} + c
Since x = a \tan(u), then u = \arctan(\frac{x}{a}). Therefore the integral becomes: \frac{1}{a} \arctan(\frac{x}{a}) + c
So, \int \frac{1}{1 + x^2} dx = \arctan(x) + c

Evaluate \int_{0}^{\frac{1}{2}} \frac{x^3}{\sqrt{1 - x^2}} dx using x = \sin(u).
Evaluate \int \frac{1}{a^2 - x^2} dx by letting x = a \sin(u).