Solved Examples

Important Theorems and Properties Recap

1. Fundamental Theorems of Calculus

• First Theorem: If $$A(x) = \int_a^x f(t)\,dt,$$ then $$A'(x) = f(x).$$
• Second Theorem: If $F'(x)=f(x)$, then $$\int_a^b f(x)\,dx = F(b)-F(a).$$

2. Core Properties of Definite Integrals

• P0 (Dummy Variable): $$\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$$
• P1 (Reversal of Limits): $$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$
• P2 (Splitting): $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$
• P3 (King's Rule): $$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$$
• P4 (Special Case of P3): $$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$$
• P5: $$\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx$$
• P6: $$\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx \quad \text{if } f(2a-x)=f(x)$$ and $$\int_0^{2a} f(x)\,dx = 0 \quad \text{if } f(2a-x)=-f(x).$$
• P7 (Even/Odd Function Property): $$\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx \quad \text{if } f(-x)=f(x),$$ $$\int_{-a}^a f(x)\,dx = 0 \quad \text{if } f(-x)=-f(x).$$

3. Integration by Parts

$$\int u\,dv = uv - \int v\,du$$ (Use the ILATE rule to choose $u$ wisely.)

4. Special Exponential Form

If an integrand can be written as $$e^x[f(x)+f'(x)],$$ then $$\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C.$$

Example 1: Basic Integration

Evaluate: $$\int \left(2x^2 + e^x + \frac{1}{x}\right) dx$$

Solution: Step 1: Split the integral into simpler parts: $$\int 2x^2\,dx + \int e^x\,dx + \int \frac{1}{x}\,dx$$

Step 2: Use standard formulas:

• $$\int 2x^2\,dx = 2\cdot \frac{x^3}{3} = \frac{2x^3}{3}$$
• $$\int e^x\,dx = e^x$$
• $$\int \frac{1}{x}\,dx = \ln|x|$$

Step 3: Combine all results and add the constant of integration.

Answer: $$\frac{2}{3}x^3 + e^x + \ln|x| + C$$

Example 2: Simplifying the Integrand

Evaluate: $$\int \frac{x^3 + 5x^2 - 4}{x^2} dx$$

Solution: Step 1: Divide each term in the numerator by $x^2$: $$\frac{x^3}{x^2} + \frac{5x^2}{x^2} - \frac{4}{x^2} = x + 5 - 4x^{-2}$$ So the integral becomes: $$\int (x + 5 - 4x^{-2})\,dx$$

Step 2: Integrate term by term:

• $$\int x\,dx = \frac{x^2}{2}$$
• $$\int 5\,dx = 5x$$
• $$\int -4x^{-2}\,dx = -4\left(\frac{x^{-1}}{-1}\right)=4x^{-1}=\frac{4}{x}$$

Step 3: Combine all terms.

Answer: $$\frac{x^2}{2} + 5x + \frac{4}{x} + C$$

Example 3: Trigonometric Identities

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

Solution: Step 1: Use the half-angle identity: $$\cos^2 x = \frac{1+\cos 2x}{2}$$ So, $$\int \cos^2 x\,dx = \int \frac{1+\cos 2x}{2}\,dx$$

Step 2: Separate the terms: $$= \frac{1}{2}\int 1\,dx + \frac{1}{2}\int \cos 2x\,dx$$

Step 3: Integrate:

• $$\frac{1}{2}\int 1\,dx = \frac{x}{2}$$
• $$\frac{1}{2}\int \cos 2x\,dx = \frac{1}{2}\cdot \frac{\sin 2x}{2} = \frac{\sin 2x}{4}$$

Answer: $$\frac{x}{2} + \frac{\sin 2x}{4} + C$$

Example 4: Integration by Substitution

Evaluate: $$\int \frac{2x}{1+x^2} dx$$

Solution: Step 1: Let $$t = 1+x^2$$ Then $$dt = 2x\,dx$$

Step 2: Substitute into the integral: $$\int \frac{2x}{1+x^2} dx = \int \frac{dt}{t}$$

Step 3: Integrate: $$\int \frac{dt}{t} = \ln|t| + C$$

Step 4: Substitute back $t=1+x^2$: $$\ln|1+x^2| + C$$ Since $1+x^2>0$ for all real $x$, the modulus is unnecessary.

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

Example 5: Substitution with Logarithm

Evaluate: $$\int \frac{(\ln x)^2}{x} dx$$

Solution: Step 1: Let $$t = \ln x$$ Then $$dt = \frac{1}{x}dx$$

Step 2: Substitute into the integral: $$\int \frac{(\ln x)^2}{x} dx = \int t^2\,dt$$

Step 3: Integrate: $$\int t^2\,dt = \frac{t^3}{3} + C$$

Step 4: Replace $t$ by $\ln x$.

Answer: $$\frac{(\ln x)^3}{3} + C$$

Example 6: Substitution with Inverse Trigonometry

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

Solution: Step 1: Let $$t = \tan^{-1}x$$ Then $$dt = \frac{1}{1+x^2}dx$$

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

Step 3: Integrate: $$\int e^t\,dt = e^t + C$$

Step 4: Substitute back $t=\tan^{-1}x$.

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

Example 7: Odd Powers of Sine and Cosine

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

Solution: Step 1: Since the power of sine is odd, separate one $\sin x$: $$\sin^3 x = \sin^2 x\sin x = (1-\cos^2 x)\sin x$$ So the integral becomes: $$\int (1-\cos^2 x)\cos^2 x\sin x\,dx$$

Step 2: Let $$t = \cos x$$ Then $$dt = -\sin x\,dx \implies -dt = \sin x\,dx$$

Step 3: Substitute: $$\int (1-t^2)t^2(-dt) = -\int (t^2-t^4)dt = \int (t^4-t^2)dt$$

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

Step 5: Put back $t=\cos x$.

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

Example 8: Product to Sum Formulas

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

Solution: Step 1: Use the identity $$2\sin A\cos B = \sin(A+B) + \sin(A-B).$$ Here $A=3x$ and $B=4x$. Thus, $$2\sin 3x\cos 4x = \sin 7x + \sin(-x) = \sin 7x - \sin x$$

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

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

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

Example 9: Completing the Square

Evaluate: $$\int \frac{dx}{x^2 - 6x + 13}$$

Solution: Step 1: Complete the square: $$x^2 - 6x + 13 = x^2 - 6x + 9 - 9 + 13 = (x-3)^2 + 4$$ So, $$x^2 - 6x + 13 = (x-3)^2 + 2^2$$

Step 2: Rewrite the integral: $$\int \frac{dx}{(x-3)^2 + 2^2}$$

Step 3: Use the standard result: $$\int \frac{dx}{X^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{X}{a}\right) + C$$ Here $X=x-3$ and $a=2$.

Answer: $$\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right) + C$$

Example 10: Completing the Square Inside a Root

Evaluate: $$\int \frac{dx}{\sqrt{5-4x-x^2}}$$

Solution: Step 1: Rewrite the expression inside the root: $$5-4x-x^2 = 5-(x^2+4x)$$ Now complete the square inside the bracket: $$x^2+4x = x^2+4x+4-4 = (x+2)^2 - 4$$ Therefore, $$5-4x-x^2 = 5-[(x+2)^2-4] = 9-(x+2)^2 = 3^2-(x+2)^2$$

Step 2: So the integral becomes: $$\int \frac{dx}{\sqrt{3^2-(x+2)^2}}$$

Step 3: Use the standard formula: $$\int \frac{dx}{\sqrt{a^2-X^2}} = \sin^{-1}\left(\frac{X}{a}\right)+C$$ with $X=x+2$ and $a=3$.

Answer: $$\sin^{-1}\left(\frac{x+2}{3}\right) + C$$

Example 11: Linear / Quadratic Form

Evaluate: $$\int \frac{x+2}{2x^2+6x+5} dx$$

Solution: Step 1: Write the numerator in the form $$x+2 = A\frac{d}{dx}(2x^2+6x+5) + B$$ Since $$\frac{d}{dx}(2x^2+6x+5)=4x+6,$$ we need $$x+2 = A(4x+6)+B = 4Ax + (6A+B).$$

Step 2: Compare coefficients: $$4A = 1 \implies A=\frac14$$ $$6A+B = 2 \implies 6\left(\frac14\right)+B=2 \implies \frac32 + B = 2 \implies B=\frac12$$

Step 3: Split the integral: $$I = \frac14\int \frac{4x+6}{2x^2+6x+5}dx + \frac12\int \frac{dx}{2x^2+6x+5}$$

Step 4: Evaluate the first part: $$\frac14\int \frac{4x+6}{2x^2+6x+5}dx = \frac14 \ln|2x^2+6x+5|$$

Step 5: For the second part, complete the square: $$2x^2+6x+5 = 2\left(x^2+3x+\frac52\right)$$ Now, $$x^2+3x+\frac52 = x^2+3x+\frac94 - \frac94 + \frac52 = \left(x+\frac32\right)^2 + \frac14$$ Hence, $$2x^2+6x+5 = 2\left[\left(x+\frac32\right)^2 + \left(\frac12\right)^2\right]$$ So, $$\int \frac{dx}{2x^2+6x+5} = \frac12\int \frac{dx}{\left(x+\frac32\right)^2 + \left(\frac12\right)^2}$$ Using the standard formula, $$= \tan^{-1}(2x+3)$$ Thus the contribution of this part is $$\frac12\tan^{-1}(2x+3).$$

Answer: $$\frac14\ln|2x^2+6x+5| + \frac12\tan^{-1}(2x+3) + C$$

Example 12: Partial Fractions (Distinct Linear Factors)

Evaluate: $$\int \frac{dx}{(x+1)(x+2)}$$

Solution: Step 1: Assume $$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$

Step 2: Multiply by $(x+1)(x+2)$: $$1 = A(x+2) + B(x+1)$$

Step 3: Find constants:

• Put $x=-1$: $$1 = A(1) \implies A=1$$
• Put $x=-2$: $$1 = B(-1) \implies B=-1$$

Step 4: Rewrite and integrate: $$\int \left(\frac{1}{x+1} - \frac{1}{x+2}\right)dx$$ $$= \ln|x+1| - \ln|x+2| + C$$ $$= \ln\left|\frac{x+1}{x+2}\right| + C$$

Answer: $$\ln\left|\frac{x+1}{x+2}\right| + C$$

Example 13: Partial Fractions (Improper Fraction)

Evaluate: $$\int \frac{x^2+1}{x^2-5x+6} dx$$

Solution: Step 1: Since the degree of the numerator equals the degree of the denominator, first divide: $$\frac{x^2+1}{x^2-5x+6} = 1 + \frac{5x-5}{x^2-5x+6}$$

Step 2: Factor the denominator: $$x^2-5x+6 = (x-2)(x-3)$$ So we resolve $$\frac{5x-5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$$

Step 3: Multiply through: $$5x-5 = A(x-3) + B(x-2)$$ Now substitute convenient values:

• Put $x=2$: $$10-5 = A(-1) \implies 5=-A \implies A=-5$$
• Put $x=3$: $$15-5 = B(1) \implies B=10$$

Step 4: Integrate: $$\int \left(1 - \frac{5}{x-2} + \frac{10}{x-3}\right)dx$$ $$= x - 5\ln|x-2| + 10\ln|x-3| + C$$

Answer: $$x - 5\ln|x-2| + 10\ln|x-3| + C$$

Example 14: Partial Fractions (Repeated Linear Factor)

Evaluate: $$\int \frac{x}{(x-1)^2(x+2)} dx$$

Solution: Step 1: Write the decomposition: $$\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$$

Step 2: Multiply by the common denominator: $$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$$

Step 3: Find constants.

• Put $x=1$: $$1 = 3B \implies B=\frac13$$
• Put $x=-2$: $$-2 = 9C \implies C=-\frac29$$
• Compare the coefficients of $x^2$: $$0 = A + C \implies A = -C = \frac29$$

Step 4: Integrate: $$\int \left(\frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2}\right)dx$$ Now, $$\int \frac{1}{(x-1)^2}dx = \int (x-1)^{-2}dx = -\frac{1}{x-1}$$ So, $$= \frac29\ln|x-1| - \frac{1}{3(x-1)} - \frac29\ln|x+2| + C$$ $$= \frac29\ln\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C$$

Answer: $$\frac29\ln\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C$$

Example 15: Integration by Parts (Basic)

Evaluate: $$\int x\sin x\,dx$$

Solution: Step 1: By ILATE, choose $$u=x, \qquad dv=\sin x\,dx$$ Then $$du=dx, \qquad v=-\cos x$$

Step 2: Apply the formula $$\int u\,dv = uv - \int v\,du$$ So, $$\int x\sin x\,dx = x(-\cos x) - \int (-\cos x)dx$$ $$= -x\cos x + \int \cos x\,dx$$ $$= -x\cos x + \sin x + C$$

Answer: $$-x\cos x + \sin x + C$$

Example 16: Integration by Parts (Logarithmic)

Evaluate: $$\int \ln x\,dx$$

Solution: Step 1: Write it as $$\int \ln x \cdot 1\,dx$$ Choose $$u=\ln x, \qquad dv=dx$$ Then $$du=\frac{1}{x}dx, \qquad v=x$$

Step 2: Apply integration by parts: $$\int \ln x\,dx = x\ln x - \int x\cdot \frac{1}{x}dx$$ $$= x\ln x - \int 1\,dx$$ $$= x\ln x - x + C$$

Answer: $$x(\ln x - 1) + C$$

Example 17: Integration by Parts (Inverse Trigonometric)

Evaluate: $$\int \sin^{-1}x\,dx$$

Solution: Step 1: Write it as $$\int \sin^{-1}x \cdot 1\,dx$$ Choose $$u=\sin^{-1}x, \qquad dv=dx$$ Then $$du=\frac{1}{\sqrt{1-x^2}}dx, \qquad v=x$$

Step 2: Apply integration by parts: $$\int \sin^{-1}x\,dx = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}}dx$$

Step 3: Evaluate the remaining integral by substitution. Let $$t = 1-x^2 \implies dt = -2x\,dx \implies x\,dx = -\frac{dt}{2}$$ Then, $$\int \frac{x}{\sqrt{1-x^2}}dx = \int \frac{-dt/2}{\sqrt t} = -\frac12\int t^{-1/2}dt = -\sqrt t$$ Thus, $$-\int \frac{x}{\sqrt{1-x^2}}dx = -(-\sqrt t)=\sqrt t$$

Step 4: Substitute back $t=1-x^2$.

Answer: $$x\sin^{-1}x + \sqrt{1-x^2} + C$$

Example 18: Cyclic Integration by Parts

Evaluate: $$I = \int e^x\sin x\,dx$$

Solution: Step 1: First application of integration by parts. Choose $$u=\sin x, \qquad dv=e^x dx$$ Then $$du=\cos x\,dx, \qquad v=e^x$$ So, $$I = e^x\sin x - \int e^x\cos x\,dx$$

Step 2: Apply integration by parts again to $$\int e^x\cos x\,dx$$ Choose $$u=\cos x, \qquad dv=e^x dx$$ Then $$du=-\sin x\,dx, \qquad v=e^x$$ Hence, $$\int e^x\cos x\,dx = e^x\cos x - \int e^x(-\sin x)dx = e^x\cos x + I$$

Step 3: Substitute back into the first equation: $$I = e^x\sin x - (e^x\cos x + I)$$ $$I = e^x\sin x - e^x\cos x - I$$ $$2I = e^x(\sin x - \cos x)$$

Step 4: Solve for $I$.

Answer: $$\frac{e^x}{2}(\sin x - \cos x) + C$$

Example 19: Special Form $\int e^x[f(x)+f'(x)]dx$

Evaluate: $$\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)dx$$

Solution: Step 1: Compare the integrand with the pattern $$e^x[f(x)+f'(x)]$$ Let $$f(x)=\frac{1}{x}$$ Then $$f'(x)=-\frac{1}{x^2}$$

Step 2: Therefore, $$\frac{1}{x} - \frac{1}{x^2} = f(x)+f'(x)$$ So the integral is exactly of the required form.

Step 3: Use the theorem: $$\int e^x[f(x)+f'(x)]dx = e^x f(x) + C$$

Answer: $$\frac{e^x}{x} + C$$

Example 20: Manipulating into the Special Form

Evaluate: $$\int \frac{xe^x}{(1+x)^2} dx$$

Solution: Step 1: Rewrite the algebraic part: $$\frac{x}{(x+1)^2} = \frac{x+1-1}{(x+1)^2} = \frac{1}{x+1} - \frac{1}{(x+1)^2}$$ So the integral becomes: $$\int e^x\left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right)dx$$

Step 2: Let $$f(x)=\frac{1}{x+1}$$ Then $$f'(x)=-\frac{1}{(x+1)^2}$$ Thus the integrand is $$e^x[f(x)+f'(x)]$$

Step 3: Apply the theorem.

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

Example 21: Definite Integral Basics

Evaluate: $$\int_0^1 \frac{dx}{1+x^2}$$

Solution: Step 1: Use the standard anti-derivative: $$\int \frac{dx}{1+x^2} = \tan^{-1}x$$

Step 2: Apply the limits: $$\int_0^1 \frac{dx}{1+x^2} = [\tan^{-1}x]_0^1$$ $$= \tan^{-1}(1) - \tan^{-1}(0)$$ $$= \frac{\pi}{4} - 0$$

Answer: $$\frac{\pi}{4}$$

Example 22: Definite Integral with Substitution

Evaluate: $$\int_0^{\pi/2} \sin^3 x\cos x\,dx$$

Solution: Step 1: Let $$t = \sin x \implies dt = \cos x\,dx$$

Step 2: Change the limits:

• When $x=0$, $t=\sin 0=0$
• When $x=\pi/2$, $t=\sin(\pi/2)=1$

Step 3: Rewrite the integral: $$\int_0^1 t^3\,dt$$

Step 4: Evaluate: $$\left[\frac{t^4}{4}\right]_0^1 = \frac14$$

Answer: $$\frac{1}{4}$$

Example 23: Definite Integral Splitting Property (P2)

Evaluate: $$\int_{-1}^2 |x^3-x|\,dx$$

Solution: Step 1: Find the zeros of the expression inside modulus: $$x^3-x = x(x-1)(x+1)=0$$ So the critical points are $x=-1,0,1$.

Step 2: Determine signs:

• On $(-1,0)$, $x^3-x>0$
• On $(0,1)$, $x^3-x<0$
• On $(1,2)$, $x^3-x>0$

Step 3: Split the integral: $$\int_{-1}^2 |x^3-x|dx = \int_{-1}^0 (x^3-x)dx + \int_0^1 -(x^3-x)dx + \int_1^2 (x^3-x)dx$$

Step 4: Use $$\int (x^3-x)dx = \frac{x^4}{4} - \frac{x^2}{2}$$ Then evaluate: $$= \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 - \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_0^1 + \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_1^2$$ This gives $$\frac14 + \frac14 + \frac94 = \frac{11}{4}$$

Answer: $$\frac{11}{4}$$

Example 24: King's Rule (P4)

Evaluate: $$I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$

Solution: Step 1: Apply Property P4 with $x \to \frac{\pi}{2}-x$: $$I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$

Step 2: Add this to the original integral: $$2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_0^{\pi/2} 1\,dx$$

Step 3: Evaluate: $$2I = \left[x\right]_0^{\pi/2} = \frac{\pi}{2}$$ Hence, $$I=\frac{\pi}{4}$$

Answer: $$\frac{\pi}{4}$$

Example 25: Removing '$x$' using P4

Evaluate: $$I = \int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} dx$$

Solution: Step 1: Apply Property P4 with $x \to \pi-x$: $$I = \int_0^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}dx$$ Using $$\sin(\pi-x)=\sin x, \qquad \cos^2(\pi-x)=\cos^2 x,$$ we get $$I = \int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}dx$$

Step 2: Add this to the original expression: $$2I = \int_0^{\pi} \frac{\pi\sin x}{1+\cos^2 x}dx$$ $$= \pi \int_0^{\pi} \frac{\sin x}{1+\cos^2 x}dx$$

Step 3: Substitute $$t = \cos x, \qquad dt=-\sin x\,dx$$ Change limits:

• $x=0 \implies t=1$
• $x=\pi \implies t=-1$ So, $$2I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2}$$

Step 4: Integrate: $$2I = \pi [\tan^{-1}t]_{-1}^1 = \pi\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right)=\frac{\pi^2}{2}$$ Thus, $$I=\frac{\pi^2}{4}$$

Answer: $$\frac{\pi^2}{4}$$

Example 26: Even and Odd Functions (P7)

Evaluate: $$\int_{-1}^1 \sin^5 x\cos^4 x\,dx$$

Solution: Step 1: Let $$f(x)=\sin^5 x\cos^4 x$$ Then $$f(-x)=\sin^5(-x)\cos^4(-x)=(-\sin x)^5(\cos x)^4=-f(x)$$ So $f(x)$ is odd.

Step 2: For an odd function over symmetric limits, $$\int_{-a}^a f(x)dx=0$$ Hence, $$\int_{-1}^1 \sin^5 x\cos^4 x\,dx=0$$

Answer: $$0$$

Example 27: Applying Property 6

Evaluate: $$\int_0^{2\pi} \cos^5 x\,dx$$

Solution: Step 1: Let $$f(x)=\cos^5 x$$ For the interval $[0,2\pi]$, the midpoint is $\pi$. Check: $$f(2\pi-x)=\cos^5(2\pi-x)=\cos^5 x = f(x)$$ Hence, $$\int_0^{2\pi} \cos^5 x\,dx = 2\int_0^{\pi} \cos^5 x\,dx$$

Step 2: Now consider the interval $[0,\pi]$, whose midpoint is $\pi/2$. Check: $$f(\pi-x)=\cos^5(\pi-x)=(-\cos x)^5=-\cos^5 x=-f(x)$$ Therefore, $$\int_0^{\pi} \cos^5 x\,dx = 0$$ So the original integral is also $0$.

Answer: $$0$$

JEE Main Standard

Example 28: Advanced Substitution

Evaluate: $$\int \frac{dx}{x^2(x^4+1)^{3/4}}$$

Solution: Step 1: Rewrite $(x^4+1)^{3/4}$ by factoring out $x^4$ inside the bracket: $$x^4+1 = x^4\left(1+x^{-4}\right)$$ Hence, $$(x^4+1)^{3/4} = (x^4)^{3/4}(1+x^{-4})^{3/4} = x^3(1+x^{-4})^{3/4}$$ (using the standard simplification used in such substitution problems).

Step 2: Therefore, $$\int \frac{dx}{x^2(x^4+1)^{3/4}} = \int \frac{x^{-5}}{(1+x^{-4})^{3/4}} dx$$

Step 3: Let $$t = 1+x^{-4}$$ Then $$dt = -4x^{-5}dx \implies x^{-5}dx = -\frac{dt}{4}$$

Step 4: Substitute: $$\int \frac{x^{-5}}{(1+x^{-4})^{3/4}} dx = -\frac14 \int t^{-3/4}dt$$

Step 5: Integrate: $$-\frac14 \int t^{-3/4}dt = -\frac14 \cdot \frac{t^{1/4}}{1/4} + C = -t^{1/4} + C$$

Step 6: Put back $t=1+x^{-4}$.

Answer: $$-(1+x^{-4})^{1/4} + C$$

Equivalent form (for $x>0$): $$-\frac{(x^4+1)^{1/4}}{x} + C$$

Example 29: Algebraic Manipulation

Evaluate: $$\int \frac{x^2-1}{x^4+3x^2+1} dx$$

Solution: Step 1: Divide numerator and denominator by $x^2$: $$\int \frac{1-\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}} dx$$

Step 2: Let $$t = x + \frac{1}{x}$$ Then, $$dt = \left(1-\frac{1}{x^2}\right)dx$$ which matches the numerator perfectly.

Step 3: Express the denominator in terms of $t$: $$t^2 = \left(x+\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$ So, $$x^2 + \frac{1}{x^2} = t^2 - 2$$ Hence the denominator becomes: $$x^2 + 3 + \frac{1}{x^2} = (t^2-2)+3 = t^2+1$$

Step 4: Substitute: $$\int \frac{dt}{t^2+1} = \tan^{-1}t + C$$

Step 5: Replace $t$ by $x+\frac{1}{x}$.

Answer: $$\tan^{-1}\left(x+\frac{1}{x}\right) + C$$

Equivalent form: $$\tan^{-1}\left(\frac{x^2+1}{x}\right) + C$$

Example 30: Definite Integral Trick

Evaluate: $$I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} dx$$

Solution: Step 1: Use the symmetry identity $$\int_{-a}^{a} f(x)dx = \int_0^a [f(x)+f(-x)]dx$$ Here, $$f(x)=\frac{\sin^2 x}{1+2^x}$$

Step 2: Compute $f(x)+f(-x)$: $$f(-x)=\frac{\sin^2(-x)}{1+2^{-x}} = \frac{\sin^2 x}{1+2^{-x}}$$ Therefore, $$f(x)+f(-x)=\sin^2 x\left(\frac{1}{1+2^x} + \frac{1}{1+2^{-x}}\right)$$ Now, $$\frac{1}{1+2^{-x}} = \frac{2^x}{1+2^x}$$ So, $$f(x)+f(-x)=\sin^2 x\left(\frac{1}{1+2^x} + \frac{2^x}{1+2^x}\right)=\sin^2 x$$

Step 3: Hence, $$I = \int_0^{\pi/2} \sin^2 x\,dx$$ Use $$\sin^2 x = \frac{1-\cos 2x}{2}$$ So, $$I=\int_0^{\pi/2} \frac{1-\cos 2x}{2}dx = \frac12\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2}$$ Since $\sin \pi = 0$ and $\sin 0 = 0$, $$I = \frac12\cdot \frac{\pi}{2} = \frac{\pi}{4}$$

Answer: $$\frac{\pi}{4}$$