Math Notes For Class 12 Chapter 3 Ex 3.2 Solved & Quiz

Question 1 (Part i): Evaluate the indefinite integral: $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx$

Solution:
Express square roots as fractional exponents: $$\int \left(x^{1/2} + x^{-1/2}\right) dx$$ Apply the linearity property of integration to separate terms: $$= \int x^{1/2} dx + \int x^{-1/2} dx$$ Apply the Power Rule of Integration ($\int x^n dx = \frac{x^{n+1}}{n+1} + C$): $$= \frac{x^{1/2 + 1}}{\frac{1}{2} + 1} + \frac{x^{-1/2 + 1}}{-\frac{1}{2} + 1} + C$$ $$= \frac{x^{3/2}}{\frac{3}{2}} + \frac{x^{1/2}}{\frac{1}{2}} + C$$ Simplify fractions: $$= \frac{2}{3}x^{3/2} + 2x^{1/2} + C = \frac{2}{3}x^{3/2} + 2\sqrt{x} + C$$
Answer: $$\frac{2}{3}x^{3/2} + 2\sqrt{x} + C$$

Question 1 (Part ii): Evaluate the integral: $\int \frac{(1 - \sqrt{x})^2}{\sqrt{x}} dx$

Solution:
Expand the numerator using algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$: $$(1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + (\sqrt{x})^2 = 1 - 2\sqrt{x} + x$$ Divide each term in the numerator by $\sqrt{x} = x^{1/2}$: $$\frac{1 - 2x^{1/2} + x}{x^{1/2}} = \frac{1}{x^{1/2}} - \frac{2x^{1/2}}{x^{1/2}} + \frac{x}{x^{1/2}} = x^{-1/2} - 2 + x^{1/2}$$ Integrate term-by-term: $$\int \left(x^{-1/2} - 2 + x^{1/2}\right) dx = \int x^{-1/2} dx - 2 \int 1 dx + \int x^{1/2} dx$$ $$= \frac{x^{1/2}}{\frac{1}{2}} - 2x + \frac{x^{3/2}}{\frac{3}{2}} + C$$ $$= 2\sqrt{x} - 2x + \frac{2}{3}x^{3/2} + C$$
Answer: $$2\sqrt{x} - 2x + \frac{2}{3}x^{3/2} + C$$

Question 2 (Part i): Evaluate the trigonometric integral: $\int \sin^2 x \, dx$

Solution:
Recall the half-angle trigonometric identity: $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
Substitute this into the integrand: $$\int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} dx$$ Separate into two distinct integrals: $$= \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos(2x) dx$$ Integrate each component: $$\int 1 dx = x, \quad \int \cos(2x) dx = \frac{\sin(2x)}{2}$$ Combining results: $$= \frac{1}{2} x - \frac{1}{2} \left(\frac{\sin(2x)}{2}\right) + C = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$
Answer: $$\frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$

Question 2 (Part ii): Evaluate the integral: $\int \frac{\cos x}{\sqrt{\sin x}} \, dx$

Solution:
Use the substitution method. Let $u = \sin x$.
Differentiating $u$ with respect to $x$ gives: $$\frac{du}{dx} = \cos x \implies du = \cos x \, dx$$ Substitute $u$ and $du$ into the original integral: $$\int \frac{\cos x \, dx}{\sqrt{\sin x}} = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} du$$ Apply the Power Rule: $$= \frac{u^{-1/2 + 1}}{-\frac{1}{2} + 1} + C = \frac{u^{1/2}}{\frac{1}{2}} + C = 2\sqrt{u} + C$$ Substitute back $u = \sin x$: $$= 2\sqrt{\sin x} + C$$
Answer: $$2\sqrt{\sin x} + C$$

Question 3 (Part i): Evaluate using variable substitution: $\int x \sqrt{x^2 + 5} \, dx$

Solution:
Let $u = x^2 + 5$.
Differentiating yields $\frac{du}{dx} = 2x \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$.
Substitute into the integral: $$\int \sqrt{x^2 + 5} \cdot (x dx) = \int \sqrt{u} \cdot \left(\frac{du}{2}\right) = \frac{1}{2} \int u^{1/2} du$$ Apply Power Rule: $$= \frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$ Substitute back $u = x^2 + 5$: $$= \frac{1}{3} (x^2 + 5)^{3/2} + C$$
Answer: $$\frac{1}{3}(x^2 + 5)^{3/2} + C$$

Question 3 (Part ii): Evaluate the integral: $\int \frac{e^x}{1 + e^x} \, dx$

Solution:
Observe that the numerator is the exact derivative of the denominator: $$\frac{d}{dx}(1 + e^x) = e^x$$ Let $u = 1 + e^x$, then $du = e^x dx$.
Using the logarithmic derivative rule $\int \frac{g'(x)}{g(x)} dx = \ln|g(x)| + C$: $$\int \frac{e^x dx}{1 + e^x} = \int \frac{du}{u} = \ln|u| + C$$ Substitute $u = 1 + e^x$ back (since $1 + e^x > 0$ for all real $x$, absolute value bars can be changed to parentheses): $$= \ln(1 + e^x) + C$$
Answer: $$\ln(1 + e^x) + C$$

Question 4 (Part i): Evaluate the logarithmic integrand: $\int \frac{1}{x \ln x} \, dx$

Solution:
Rewrite the integrand to highlight the ratio of a function and its derivative: $$\int \frac{1}{x \ln x} dx = \int \frac{\frac{1}{x}}{\ln x} dx$$ Let $u = \ln x$. Differential $du = \frac{1}{x} dx$.
Substitute into the integral: $$\int \frac{du}{u} = \ln|u| + C$$ Substitute back $u = \ln x$: $$= \ln|\ln x| + C$$
Answer: $$\ln|\ln x| + C$$

Question 4 (Part ii): Evaluate the exponential integral: $\int a^{2x} \, dx \quad (a > 0, a \neq 1)$

Solution:
Recall the formula for integrating exponential functions of base $a$: $\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$.
Alternatively, substitute $u = 2x \implies du = 2dx \implies dx = \frac{du}{2}$: $$\int a^{2x} dx = \int a^u \left(\frac{du}{2}\right) = \frac{1}{2} \int a^u du = \frac{1}{2} \cdot \frac{a^u}{\ln a} + C$$ Substitute $u = 2x$ back: $$= \frac{a^{2x}}{2 \ln a} + C$$
Answer: $$\frac{a^{2x}}{2 \ln a} + C$$

Question 5 (Part i): Evaluate: $\int \frac{x+2}{x^2 + 4x + 7} \, dx$

Solution:
Let $u = x^2 + 4x + 7$.
Compute derivative: $$\frac{du}{dx} = 2x + 4 = 2(x + 2) \implies (x + 2) dx = \frac{du}{2}$$ Substitute this into the expression: $$\int \frac{x+2}{x^2 + 4x + 7} dx = \int \frac{\frac{du}{2}}{u} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C$$ Substitute $u = x^2 + 4x + 7$ back (note $x^2 + 4x + 7 = (x+2)^2 + 3 > 0$ for all real $x$): $$= \frac{1}{2} \ln(x^2 + 4x + 7) + C$$
Answer: $$\frac{1}{2} \ln(x^2 + 4x + 7) + C$$

Question 5 (Part ii): Evaluate: $\int \frac{\sec^2 x}{\tan x} \, dx$

Solution:
Notice that $\frac{d}{dx}(\tan x) = \sec^2 x$.
Let $u = \tan x \implies du = \sec^2 x \, dx$.
Substitute into the integrand: $$\int \frac{\sec^2 x}{\tan x} dx = \int \frac{du}{u} = \ln|u| + C$$ Substitute $u = \tan x$ back: $$= \ln|\tan x| + C$$
Answer: $$\ln|\tan x| + C$$

Question 6: Find the function $f(x)$ if its derivative is given by $f'(x) = 3x^2 - 4x + 5$ and $f(1) = 6$.

Solution:
Integrate $f'(x)$ to find the general expression for $f(x)$: $$f(x) = \int f'(x) dx = \int (3x^2 - 4x + 5) dx$$ Apply power rule term-by-term: $$f(x) = 3 \left(\frac{x^3}{3}\right) - 4 \left(\frac{x^2}{2}\right) + 5x + C$$ $$f(x) = x^3 - 2x^2 + 5x + C$$ Use the initial condition $f(1) = 6$ to determine constant $C$: $$6 = (1)^3 - 2(1)^2 + 5(1) + C$$ $$6 = 1 - 2 + 5 + C \implies 6 = 4 + C \implies C = 2$$ Substitute $C = 2$ back into the function: $$f(x) = x^3 - 2x^2 + 5x + 2$$
Answer: $$f(x) = x^3 - 2x^2 + 5x + 2$$