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

Integration by Substitution: Indefinite Integrals & Inverse Trigonometric Functions

Revisiting Integration by Substitution

• 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 $f'(g(x))g'(x)$.

Integrals Involving Powers of Trigonometric Functions

• 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 $\tan^2(x) + 1 = \sec^2(x)$.

Example: Integrating $\sin^3(x) \cos^2(x)$

• 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: $-\frac{\cos^3(x)}{3} + \frac{\cos^5(x)}{5} + c$

Integrals Involving $\sqrt{a^2 - x^2}$

• 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, $\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2(u)} = \sqrt{a^2(1 - \sin^2(u))} = a\sqrt{\cos^2(u)} = a|\cos(u)|$

Example: Definite Integral with $\sqrt{1 - x^2}$

• 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: $\int<em>{0}^{\frac{\pi}{2}} \frac{\sin^3(u)}{\sqrt{1 - \sin^2(u)}} \cos(u) du = \int</em>{0}^{\frac{\pi}{2}} \sin^3(u) \cos(u) \frac{1}{|\cos(u)|} du$
• 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.

Integrals Involving $a^2 + x^2$

• 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, $\int \frac{1}{1 + x^2} dx = \arctan(x) + c$

Suggested Exercises

• 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 $x = a \sin(u)$.