Integrals Exercise 7.3: Integrating Trigonometric Functions

The Power of Trigonometric Identities in Integration

Many trigonometric functions, especially those involving powers (like $\sin^2 x$, $\cos^3 x$) or products (like $\sin(ax)\cos(bx)$), cannot be integrated directly using basic formulas. For instance, there's no direct formula for $\int \sin^2 x \, dx$ or $\int \sin x \cos x \, dx$ in terms of elementary functions that don't involve other trig functions. This is where trigonometric identities come to our rescue. The core idea is to transform the integrand (the function being integrated) into an equivalent form that consists of sums or differences of simpler trigonometric functions, each of which can be integrated using standard formulas.

For example, if we need to integrate $\sin^2 x$, we recall the double-angle identity: $\cos(2x) = 1 - 2\sin^2 x$. Rearranging this, we get $\sin^2 x = \frac{1 - \cos(2x)}{2}$. Now, $\int \frac{1 - \cos(2x)}{2} \, dx$ is straightforward: $\frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(2x) \, dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$. Similarly, product-to-sum identities are crucial for integrals like $\int \sin(Ax)\cos(Bx) \, dx$. The ability to recognize which identity to apply is key to mastering this section.

Essential Trigonometric Identities for Integration

Power Reduction for Sine

$\sin^2 x = \frac{1 - \cos(2x)}{2}$. Used to reduce the power of $\sin x$ to a form integrable in terms of $\cos(2x)$.

Power Reduction for Cosine

$\cos^2 x = \frac{1 + \cos(2x)}{2}$. Used to reduce the power of $\cos x$ to a form integrable in terms of $\cos(2x)$.

Triple Angle for Sine

$\sin(3x) = 3\sin x - 4\sin^3 x \implies \sin^3 x = \frac{3\sin x - \sin(3x)}{4}$. Useful for integrating $\sin^3 x$.

Triple Angle for Cosine

$\cos(3x) = 4\cos^3 x - 3\cos x \implies \cos^3 x = \frac{\cos(3x) + 3\cos x}{4}$. Useful for integrating $\cos^3 x$.

Product-to-Sum (Sine & Cosine)

$2\sin A \cos B = \sin(A+B) + \sin(A-B)$. Transforms products into sums, making integration easier.

Product-to-Sum (Sine & Sine)

$2\sin A \sin B = \cos(A-B) - \cos(A+B)$. Transforms products into differences.

Product-to-Sum (Cosine & Cosine)

$2\cos A \cos B = \cos(A+B) + \cos(A-B)$. Transforms products into sums.

Tangent Identity

$\tan^2 x = \sec^2 x - 1$. Since $\int \sec^2 x \, dx = \tan x + C$ and $\int 1 \, dx = x + C$, this is very useful.

Step-by-Step Worked Examples

• Example 1: Integrate $\int \sin^4 x \, dx$ Step 1: Rewrite $\sin^4 x$ using power reduction. We know $\sin^2 x = \frac{1 - \cos(2x)}{2}$. $\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}$ Step 2: Apply power reduction again for $\cos^2(2x)$. We know $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$. Here, $\theta = 2x$. $\cos^2(2x) = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos(4x)}{2}$ Step 3: Substitute back and simplify the integrand. $\sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4} = \frac{2 - 4\cos(2x) + 1 + \cos(4x)}{8} = \frac{3 - 4\cos(2x) + \cos(4x)}{8}$ Step 4: Integrate term by term. $\int \sin^4 x \, dx = \int \frac{3 - 4\cos(2x) + \cos(4x)}{8} \, dx$ $= \frac{1}{8} \left( \int 3 \, dx - \int 4\cos(2x) \, dx + \int \cos(4x) \, dx \right)$ $= \frac{1}{8} \left( 3x - 4\frac{\sin(2x)}{2} + \frac{\sin(4x)}{4} \right) + C$ Final Answer: $\frac{3x}{8} - \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C$
• Example 2: Integrate $\int \sin(5x)\cos(3x) \, dx$ Step 1: Use the product-to-sum identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$. Here $A=5x$ and $B=3x$. So, $\sin(5x)\cos(3x) = \frac{1}{2} [\sin(5x+3x) + \sin(5x-3x)]$ $= \frac{1}{2} [\sin(8x) + \sin(2x)]$ Step 2: Integrate the simplified expression term by term. $\int \sin(5x)\cos(3x) \, dx = \int \frac{1}{2} [\sin(8x) + \sin(2x)] \, dx$ $= \frac{1}{2} \left( \int \sin(8x) \, dx + \int \sin(2x) \, dx \right)$ $= \frac{1}{2} \left( -\frac{\cos(8x)}{8} - \frac{\cos(2x)}{2} \right) + C$ Final Answer: $-\frac{\cos(8x)}{16} - \frac{\cos(2x)}{4} + C$
• Example 3: Integrate $\int \frac{\sin x}{1 + \sin x} \, dx$ Step 1: Rationalize the denominator by multiplying by the conjugate. $\int \frac{\sin x}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x} \, dx = \int \frac{\sin x - \sin^2 x}{1 - \sin^2 x} \, dx$ Step 2: Use the identity $1 - \sin^2 x = \cos^2 x$. $= \int \frac{\sin x - \sin^2 x}{\cos^2 x} \, dx$ Step 3: Split the fraction into integrable terms. $= \int \left( \frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \right) \, dx$ $= \int \left( \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} - \tan^2 x \right) \, dx$ $= \int (\tan x \sec x - \tan^2 x) \, dx$ Step 4: Use the identity $\tan^2 x = \sec^2 x - 1$ for the second term. $= \int (\tan x \sec x - (\sec^2 x - 1)) \, dx$ $= \int (\tan x \sec x - \sec^2 x + 1) \, dx$ Step 5: Integrate term by term. We know $\int \sec x \tan x \, dx = \sec x + C$, $\int \sec^2 x \, dx = \tan x + C$, and $\int 1 \, dx = x + C$. Final Answer: $\sec x - \tan x + x + C$

Exam Tips for Trigonometric Integrals (Ex 7.3)

1. Memorize Identities: The most crucial step is to have the key trigonometric identities at your fingertips: power reduction formulas for $\sin^2 x$ and $\cos^2 x$, triple angle formulas for $\sin^3 x$ and $\cos^3 x$, and all product-to-sum/sum-to-product identities. Make flashcards if needed!
2. Look for Simplification: Before applying any complex identity, check if simple substitutions or splitting the fraction can work. For example, if you have $\int \sin^3 x \cos x \, dx$, a simple substitution $u = \sin x$ ($du = \cos x \, dx$) yields $\int u^3 \, du$, which is much easier than using triple angle identities.
3. Handle Higher Powers: For $\sin^n x$ or $\cos^n x$ where $n > 2$ and $n$ is even, you'll often need to apply the power reduction formulas multiple times, as seen in Example 1. For odd powers, factor out one term (e.g., $\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x)\sin x$) and then use substitution ($u = \cos x$).
4. Practice Product-to-Sum: Integrals involving products of different sine/cosine terms ($\sin(Ax)\cos(Bx)$, etc.) are direct applications of product-to-sum identities. Pay attention to the coefficients $A$ and $B$ and the $\frac{1}{2}$ factor that often comes with these identities.
5. Don't Forget '+ C': Always add the constant of integration 'C' at the end of indefinite integrals. It's a common oversight that can cost you marks.
6. Check Your Answer: If time permits, differentiate your final answer to see if you get back the original integrand. This is a great way to catch errors.

Practice Questions with Solutions

• Q: Integrate $\int \cos(2x)\cos(4x)\cos(6x) \, dx$ A: Step 1: Use the product-to-sum identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$ for the first two terms. $2\cos(2x)\cos(4x) = \cos(2x+4x) + \cos(2x-4x) = \cos(6x) + \cos(-2x) = \cos(6x) + \cos(2x)$ So, $\cos(2x)\cos(4x) = \frac{1}{2}(\cos(6x) + \cos(2x))$ Step 2: Substitute back into the integral and expand. $\int \frac{1}{2}(\cos(6x) + \cos(2x))\cos(6x) \, dx = \frac{1}{2} \int (\cos^2(6x) + \cos(2x)\cos(6x)) \, dx$ $= \frac{1}{2} \int \cos^2(6x) \, dx + \frac{1}{2} \int \cos(2x)\cos(6x) \, dx$ Step 3: Apply power reduction for $\cos^2(6x)$ and product-to-sum for $\cos(2x)\cos(6x)$. $\cos^2(6x) = \frac{1 + \cos(12x)}{2}$ $2\cos(2x)\cos(6x) = \cos(2x+6x) + \cos(2x-6x) = \cos(8x) + \cos(-4x) = \cos(8x) + \cos(4x)$ So, $\cos(2x)\cos(6x) = \frac{1}{2}(\cos(8x) + \cos(4x))$ Step 4: Substitute and integrate term by term. $= \frac{1}{2} \int \frac{1 + \cos(12x)}{2} \, dx + \frac{1}{2} \int \frac{\cos(8x) + \cos(4x)}{2} \, dx$ $= \frac{1}{4} \int (1 + \cos(12x)) \, dx + \frac{1}{4} \int (\cos(8x) + \cos(4x)) \, dx$ $= \frac{1}{4} \left( x + \frac{\sin(12x)}{12} \right) + \frac{1}{4} \left( \frac{\sin(8x)}{8} + \frac{\sin(4x)}{4} \right) + C$ Final Answer: $\frac{x}{4} + \frac{\sin(12x)}{48} + \frac{\sin(8x)}{32} + \frac{\sin(4x)}{16} + C$
• Q: Integrate $\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx$ A: Step 1: Use the double-angle identity $\cos 2A = 2\cos^2 A - 1$ for both $\cos 2x$ and $\cos 2\alpha$. $\int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx$ $= \int \frac{2\cos^2 x - 1 - 2\cos^2 \alpha + 1}{\cos x - \cos \alpha} \, dx$ $= \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx$ Step 2: Use the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$) in the numerator. $= \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \, dx$ Step 3: Cancel out the common factor $(\cos x - \cos \alpha)$. $= \int 2(\cos x + \cos \alpha) \, dx$ Step 4: Integrate term by term. Note that $\cos \alpha$ is a constant with respect to $x$. $= 2 \int \cos x \, dx + 2 \int \cos \alpha \, dx$ $= 2\sin x + 2x\cos \alpha + C$ Final Answer: $2\sin x + 2x\cos \alpha + C$
• Q: Integrate $\int \frac{1}{\sin^2 x \cos^2 x} \, dx$ A: Step 1: Replace $1$ in the numerator with the identity $\sin^2 x + \cos^2 x = 1$. $\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx$ Step 2: Split the fraction into two separate terms. $= \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx$ Step 3: Simplify each term. $= \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx$ Step 4: Rewrite in terms of $\sec^2 x$ and $\csc^2 x$. $= \int (\sec^2 x + \csc^2 x) \, dx$ Step 5: Integrate term by term. $= \int \sec^2 x \, dx + \int \csc^2 x \, dx$ $= \tan x - \cot x + C$ Final Answer: $\tan x - \cot x + C$
• Q: Integrate $\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$ A: Step 1: Rewrite the denominator $1 + \sin 2x$ using identities. We know $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2\sin x \cos x$. So, $1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2$. Step 2: Substitute this into the integral. $\int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx$ Step 3: Use substitution. Let $u = \sin x + \cos x$. Then $du = (\cos x - \sin x) \, dx$. Step 4: Substitute $u$ and $du$ into the integral. $= \int \frac{1}{u^2} \, du = \int u^{-2} \, du$ Step 5: Integrate with respect to $u$. $= \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$ Step 6: Substitute back $u = \sin x + \cos x$. Final Answer: $-\frac{1}{\sin x + \cos x} + C$

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