Methods of Integration: Substitution and Trigonometric Identities

Integration by Substitution

When an integral is not in a direct standard form, we often simplify it by changing the variable. This technique is called integration by substitution.

The key idea is to identify a part of the integrand, say $g(x)$, whose derivative $g'(x)$ also appears in the integral. Then the integral can be reduced to a simpler form.

General Rule

If $$I = \int f(g(x))\,g'(x)\,dx,$$ then we put $$t = g(x).$$ Differentiating, $$\frac{dt}{dx} = g'(x) \quad \Rightarrow \quad dt = g'(x)\,dx.$$ So the integral becomes $$I = \int f(t)\,dt.$$ After integrating with respect to $t$, we replace $t$ by $g(x)$.

Steps of the Method

1. Choose a suitable substitution $t = g(x)$.
2. Differentiate to get $dt = g'(x)dx$.
3. Rewrite the whole integral in terms of $t$.
4. Integrate with respect to $t$.
5. Substitute back in terms of $x$.

Integrals of Standard Trigonometric Functions using Substitution

Using substitution, we can derive the integrals of trigonometric functions that are not immediately obvious from standard derivatives.

1. Integral of $\tan x$: $$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx.$$ Put $t = \cos x$, so $dt = -\sin x\,dx$. Then, $$\int \tan x\,dx = -\int \frac{dt}{t} = -\ln|t| + C = -\ln|\cos x| + C = \ln|\sec x| + C.$$

2. Integral of $\cot x$: $$\int \cot x\,dx = \int \frac{\cos x}{\sin x}\,dx.$$ Put $t = \sin x$, so $dt = \cos x\,dx$. Then, $$\int \cot x\,dx = \int \frac{dt}{t} = \ln|t| + C = \ln|\sin x| + C.$$

3. Integral of $\sec x$: Multiply and divide by $(\sec x + \tan x)$: $$\int \sec x\,dx = \int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x}\,dx$$ $$= \int \frac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}\,dx.$$ Put $t = \sec x + \tan x$, so $$dt = (\sec x\tan x + \sec^2 x)dx.$$ Therefore, $$\int \sec x\,dx = \int \frac{dt}{t} = \ln|t| + C = \ln|\sec x + \tan x| + C.$$

4. Integral of $\csc x$: Multiply and divide by $(\csc x - \cot x)$: $$\int \csc x\,dx = \int \frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx$$ $$= \int \frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx.$$ Put $t = \csc x - \cot x$, so $$dt = ( -\csc x\cot x + \csc^2 x )dx.$$ Hence, $$\int \csc x\,dx = \int \frac{dt}{t} = \ln|t| + C = \ln|\csc x - \cot x| + C.$$

Integration using Trigonometric Identities

When the integrand contains powers or products of trigonometric functions, direct substitution may not be enough. In such cases, trigonometric identities help convert the integral into an easier form.

Important Identities

Half-angle formulas

$$\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}$$ These are especially useful for even powers of sine and cosine.

Pythagorean identities

$$\sin^2 x + \cos^2 x = 1$$ $$\sec^2 x - \tan^2 x = 1$$ $$\csc^2 x - \cot^2 x = 1$$ These help convert odd powers into a form suitable for substitution.

Product-to-sum formulas

$$2\sin A\cos B = \sin(A+B) + \sin(A-B)$$ $$2\cos A\sin B = \sin(A+B) - \sin(A-B)$$ $$2\cos A\cos B = \cos(A+B) + \cos(A-B)$$ $$2\sin A\sin B = \cos(A-B) - \cos(A+B)$$ These are very useful when the integrand is a product like $\sin mx\cos nx$ or $\cos mx\cos nx$.

Solved Examples

Example 1: Basic Substitution

Evaluate: $\int 2x\sin(x^2 + 1)\,dx$

Solution: Step 1: The expression $x^2+1$ is inside the sine function, and its derivative is $2x$, which is present in the integrand. Let $$t = x^2 + 1.$$

Step 2: Differentiate: $$dt = 2x\,dx.$$

Step 3: Substitute into the integral: $$\int 2x\sin(x^2+1)\,dx = \int \sin t\,dt.$$

Step 4: Integrate: $$\int \sin t\,dt = -\cos t + C.$$

Step 5: Replace $t$ by $x^2+1$: $$-\cos(x^2+1) + C.$$

Answer: $-\cos(x^2 + 1) + C$

---

Example 2: Substitution Involving Logarithms

Evaluate: $\int \frac{\ln x}{x}\,dx$

Solution: Step 1: Since the derivative of $\ln x$ is $\frac{1}{x}$, use the substitution $$t = \ln x.$$

Step 2: Then $$dt = \frac{1}{x}\,dx.$$

Step 3: Substitute: $$\int \frac{\ln x}{x}\,dx = \int t\,dt.$$

Step 4: Integrate: $$\int t\,dt = \frac{t^2}{2} + C.$$

Step 5: Put back $t = \ln x$: $$\frac{(\ln x)^2}{2} + C.$$

Answer: $\frac{(\ln x)^2}{2} + C$

---

Example 3: Substitution with Inverse Trigonometric Functions

Evaluate: $\int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx$

Solution: Step 1: The derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, which appears in the integrand. Let $$t = \tan^{-1}x.$$

Step 2: Then $$dt = \frac{1}{1+x^2}\,dx.$$

Step 3: Substitute: $$\int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx = \int e^t\,dt.$$

Step 4: Integrate: $$\int e^t\,dt = e^t + C.$$

Step 5: Replace $t$: $$e^{\tan^{-1}x} + C.$$

Answer: $e^{\tan^{-1}x} + C$

---

Example 4: Integrals of the form $f'(x)/f(x)$

Evaluate: $\int \frac{2x+3}{x^2+3x+5}\,dx$

Solution: Step 1: The denominator is $$x^2+3x+5,$$ and its derivative is $$2x+3,$$ which is exactly the numerator. Let $$t = x^2+3x+5.$$

Step 2: Then $$dt = (2x+3)\,dx.$$

Step 3: Substitute: $$\int \frac{2x+3}{x^2+3x+5}\,dx = \int \frac{1}{t}\,dt.$$

Step 4: Integrate: $$\int \frac{1}{t}\,dt = \ln|t| + C.$$

Step 5: Replace $t$: $$\ln|x^2+3x+5| + C.$$

Since $x^2+3x+5$ is always positive, the modulus is not strictly necessary, but it is standard to write it.

Answer: $\ln|x^2 + 3x + 5| + C$

---

Example 5: Integrating Even Powers of Sine/Cosine

Evaluate: $\int \sin^2(2x+5)\,dx$

Solution: Step 1: Use the half-angle identity $$\sin^2 \theta = \frac{1-\cos 2\theta}{2}.$$ Here $\theta = 2x+5$. So, $$\sin^2(2x+5) = \frac{1-\cos(4x+10)}{2}.$$

Step 2: Rewrite the integral: $$\int \sin^2(2x+5)\,dx = \frac{1}{2}\int [1-\cos(4x+10)]\,dx.$$

Step 3: Integrate term by term: $$= \frac{1}{2}\left( \int 1\,dx - \int \cos(4x+10)\,dx \right).$$

Step 4: Use $$\int \cos(ax+b)\,dx = \frac{\sin(ax+b)}{a} + C.$$ Therefore, $$= \frac{1}{2}\left( x - \frac{\sin(4x+10)}{4} \right) + C.$$

Step 5: Simplify: $$= \frac{x}{2} - \frac{\sin(4x+10)}{8} + C.$$

Answer: $\frac{x}{2} - \frac{\sin(4x + 10)}{8} + C$

---

Example 6: Integrating Odd Powers of Sine/Cosine

Evaluate: $\int \sin^3 x\cos^2 x\,dx$

Solution: Step 1: Since the power of sine is odd, separate one factor of $\sin x$: $$\sin^3 x = \sin^2 x\sin x.$$ So, $$\int \sin^3 x\cos^2 x\,dx = \int \sin^2 x\cos^2 x\sin x\,dx.$$

Step 2: Use $$\sin^2 x = 1-\cos^2 x.$$ Then, $$= \int (1-\cos^2 x)\cos^2 x\sin x\,dx.$$

Step 3: Let $$t = \cos x, \qquad dt = -\sin x\,dx.$$ So, $$-dt = \sin x\,dx.$$

Step 4: Substitute: $$\int (1-t^2)t^2(-dt) = -\int (t^2 - t^4)dt = \int (t^4 - t^2)dt.$$

Step 5: Integrate: $$\int (t^4 - t^2)dt = \frac{t^5}{5} - \frac{t^3}{3} + C.$$

Step 6: Replace $t = \cos x$: $$\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C.$$

Answer: $\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$

---

Example 7: Product to Sum Identities

Evaluate: $\int \sin 3x\cos 4x\,dx$

Solution: Step 1: Use the product-to-sum identity $$2\sin A\cos B = \sin(A+B) + \sin(A-B).$$ Take $A=3x$ and $B=4x$. Then, $$2\sin 3x\cos 4x = \sin 7x + \sin(-x) = \sin 7x - \sin x.$$

Step 2: Therefore, $$\sin 3x\cos 4x = \frac{1}{2}(\sin 7x - \sin x).$$ So the integral becomes $$\int \sin 3x\cos 4x\,dx = \frac{1}{2}\int (\sin 7x - \sin x)\,dx.$$

Step 3: Integrate term by term: $$= \frac{1}{2}\left( \int \sin 7x\,dx - \int \sin x\,dx \right).$$

Step 4: Use $$\int \sin ax\,dx = -\frac{\cos ax}{a} + C.$$ Hence, $$= \frac{1}{2}\left( -\frac{\cos 7x}{7} + \cos x \right) + C.$$

Step 5: Simplify: $$= \frac{\cos x}{2} - \frac{\cos 7x}{14} + C.$$

Answer: $\frac{\cos x}{2} - \frac{\cos 7x}{14} + C$