Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

This problem relies on the definition of a definite integral as the limit of a Riemann sum, where $\lim_{n \to \infty } \frac{1}{n} \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) = \int_0^1 f(x) \, dx$. We can write the given series as $\frac{3}{n} \sum_{r=0}^{n-1} (2 + \frac{r}{n})^2$, which allows us to identify the function $f(x) = (2+x)^2$ within the structure of the limit. Converting this into an integral with the constant factor of 3 outside results in the expression $3 \int_0^1 (2+x)^2 \, dx$. Evaluating the integral yields $\frac{(2+x)^3}{3}$, and applying the limits from 0 to 1 results in $3 \left( \frac{3^3}{3} - \frac{2^3}{3} \right)$, which simplifies to 27 - 8, yielding a final value of 19.

The integral becomes manageable by rationalizing the denominator, which involves multiplying the numerator and denominator by the conjugate $\sqrt{x+\alpha}+\sqrt{x}$. Because the denominator simplifies to $(x+\alpha)-x = \alpha$, the integrand simplifies to $\frac{1}{\alpha} (x\sqrt{x+\alpha} + x\sqrt{x})$. Integrating this term-by-term requires expressing $x$ as $(x+\alpha)-\alpha$ in the first part, which leads to the integral of $(x+\alpha)^{3/2} - \alpha(x+\alpha)^{1/2} + x^{3/2}$.

Evaluating the integral yields $\frac{1}{\alpha} \left[ \frac{2}{5}(x+\alpha)^{5/2} - \frac{2}{3}\alpha(x+\alpha)^{3/2} + \frac{2}{5}x^{5/2} \right]_0^{\alpha}$. When substituting the limits 0 and $\alpha$, the expression resolves to $\frac{\alpha^{3/2}}{15}(4\sqrt{2}+10)$. Equating this result to $\frac{16+20\sqrt{2}}{15}$ simplifies to the equation $\alpha^{3/2}(4\sqrt{2}+10) = 16+20\sqrt{2}$. Since $16+20\sqrt{2} = 2\sqrt{2}(4\sqrt{2}+10)$, it follows that $\alpha^{3/2} = 2\sqrt{2}$, confirming that $\alpha = 2$.

The denominator integral $\int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{49} x (1 + \tan^2 x) \, dx$ simplifies to $\int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{49} x \sec^2 x \, dx$. Applying integration by parts with $u = e^{-x}$ and $dv = \tan^{49} x \sec^2 x \, dx$ gives $du = -e^{-x} \, dx$ and $v = \frac{\tan^{50} x}{50}$. Substituting these into the formula $\int u \, dv = uv - \int v \, du$ yields $\left[ e^{-x} \frac{\tan^{50} x}{50} \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \frac{\tan^{50} x}{50} (-e^{-x}) \, dx$. Evaluating at the limits results in $\frac{e^{-\frac{\pi}{4}}}{50} + \frac{1}{50} \int_{0}^{\frac{\pi}{4}} e^{-x} \tan^{50} x \, dx$, which is exactly $\frac{1}{50}$ times the numerator. The ratio of the two expressions therefore simplifies to 50.

The integral $16 \int_{1}^{2} \frac{dx}{x^3 (x^{2+}2)^2}$ is best approached by identifying a substitution that transforms the denominator into a more manageable polynomial form. By observing that the derivative of the reciprocal squared term involves an $x^{-3}$ component, we can isolate this structure within the integrand to simplify the variable of integration. This strategy shifts the focus from a complex rational function to a basic polynomial integral.

Using the substitution $t = 1 + \frac{2}{x^2}$ implies that $dt = -\frac{4}{x^3} dx$. Adjusting the integration limits accordingly, we find that when $x=1$, the value of $t$ is $3$, and when $x=2$, the value of $t$ is $\frac{3}{2}$. Incorporating these changes into the expression, the integral transforms into $-4 \int_{3}^{3/2} \frac{1}{t^2} \left( \frac{t-1}{2} \right)^2 dt$. Simplifying the squared term yields $-\int_{3}^{3/2} \frac{(t-1)^2}{t^2} dt$, which reduces further to $-\int_{3}^{3/2} \left( 1 - \frac{2}{t} + \frac{1}{t^2} \right) dt$.

Evaluating this expression provides the anti-derivative $-(t - 2\ln|t| - \frac{1}{t})$ taken between the limits $3$ and $\frac{3}{2}$. Applying these limits gives the calculation $- \left[ \left( \frac{3}{2} - 2\ln\frac{3}{2} - \frac{2}{3} \right) - \left( 3 - 2\ln3 - \frac{1}{3} \right) \right]$. After resolving the arithmetic inside the brackets and simplifying the logarithmic terms, the final result is $\frac{11}{6} - \ln 4$.

To find the derivative of the function $\varphi(x)$, apply the product rule for differentiation alongside the Leibniz Integral Rule. The expression for $\varphi(x)$ is a product of $\frac{1}{\sqrt{x}}$ and the integral $\int_{\pi/4}^{x} (4\sqrt{2}\sin t - 3\varphi'(t)) \, dt$. Differentiating this product results in a term containing the integral and a term where the derivative of the integral simplifies to the integrand evaluated at the upper limit $x$.

When evaluating the derivative at $x = \frac{\pi}{4}$, the integral term itself becomes zero because the limits of integration are identical. This simplifies the derivative significantly, leaving only the Leibniz contribution. Specifically, the expression for the derivative at this point becomes:
$\varphi'\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\pi/4}} \left(4\sqrt{2} \sin\left(\frac{\pi}{4}\right) - 3\varphi'\left(\frac{\pi}{4}\right)\right)$

Using the values $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{\pi/4}} = \frac{2}{\sqrt{\pi}}$, the equation simplifies to:
$\varphi'\left(\frac{\pi}{4}\right) = \frac{2}{\sqrt{\pi}} \left(4 - 3\varphi'\left(\frac{\pi}{4}\right)\right)$
Multiplying by $\sqrt{\pi}$ and rearranging the terms to isolate $\varphi'\left(\frac{\pi}{4}\right)$ leads to:
$\sqrt{\pi} \varphi'\left(\frac{\pi}{4}\right) = 8 - 6\varphi'\left(\frac{\pi}{4}\right)$
$\left(6 + \sqrt{\pi}\right)\varphi'\left(\frac{\pi}{4}\right) = 8$
Final division yields the value:
$\varphi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$

The recursive definition $S_k(x) = C_k x + k \int_{0}^{x} S_{k-1}(t) dt$ allows us to construct each function sequentially, starting from $S_0(x) = x$. For the first constant, $C_1 = 1 - \int_{0}^{1} x dx = 1 - 0.5 = 0.5$. Substituting this into the recursive formula for $k=1$ gives $S_1(x) = 0.5 x + \int_{0}^{x} t dt = 0.5 x + 0.5 x^2$. Following the same procedure for the next constant, $C_2 = 1 - \int_{0}^{1} S_1(x) dx$, we find that $C_2 = 0$.

Using these values, we determine the expression for $S_2(x) = C_2 x + 2 \int_{0}^{x} S_1(t) dt$. Since $C_2 = 0$, this simplifies to $S_2(x) = 2 \int_{0}^{x} (0.5 t + 0.5 t^2) dt = 0.5 x^2 + \frac{x^3}{3}$. Evaluating this result at $x = 3$ yields $S_2(3) = 0.5(3^2) + \frac{3^3}{3} = 4.5 + 9 = 13.5$. Finally, given that $C_3 = 0.75$, the target value $S_2(3) + 6 C_3$ is calculated as $13.5 + 6(0.75) = 13.5 + 4.5 = 18$.

The expression $\int_0^2 f(t) dt$ represents a constant, which we can denote by $k$. This reduces the differential equation to the linear form
$f'(x) + f(x) = k$
Applying the integrating factor $e^x$ leads to the general solution $f(x) = k + C e^{-x}$, and the condition $f(0) = e^{-2}$ determines $C = e^{-2} - k$. Substituting $f(x) = k + (e^{-2} - k)e^{-x}$ into the definition $k = \int_0^2 f(t) dt$ allows us to solve for $k = e^{-2} - 1$. Calculating $f(2) = (e^{-2} - 1) + (e^{-2} - (e^{-2} - 1)) e^{-2} = 2 e^{-2} - 1$, the final value $2 f(0) - f(2)$ evaluates to $2(e^{-2}) - (2 e^{-2} - 1)$, which is 1.

Replacing the variable $x$ with $\frac{1}{x}$ in the expression $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$ yields a second relationship, $4 f(x)+5 f\left(\frac{1}{x}\right)=x+3$. By multiplying the first equation by 5 and the second by 4, you can eliminate the $f\left(\frac{1}{x}\right)$ term through subtraction, resulting in $25 f(x) - 16 f(x) = 5\left(\frac{1}{x}+3\right) - 4(x+3)$. This simplifies to $9 f(x) = \frac{5}{x} - 4x + 3$, which establishes that $f(x) = \frac{5}{9x} - \frac{4}{9}x + \frac{1}{3}$.

Integrating this expression over the interval $[1, 2]$ involves calculating $\int_{1}^{2} \left(\frac{5}{9x} - \frac{4}{9}x + \frac{1}{3}\right) dx$, which gives $\left[ \frac{5}{9} \ln x - \frac{2}{9} x^2 + \frac{1}{3} x \right]_1^2$. Evaluating this at the boundaries, we find the value $\left(\frac{5}{9} \ln 2 - \frac{8}{9} + \frac{2}{3}\right) - \left(0 - \frac{2}{9} + \frac{1}{3}\right)$, which simplifies to $\frac{5}{9} \ln 2 - \frac{1}{3}$. Multiplying this result by 18, as required by the problem, leads to $18 \left(\frac{5}{9} \ln 2 - \frac{1}{3}\right)$, resulting in $10 \log_{e} 2 - 6$.

The integral $I = \int_{0}^{1} \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$ relies on the definite integral property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$. By applying this symmetry with the transformation $x \to 1-x$, the exponential factor in the denominator converts to $\frac{1+e^{2-4x}}{e^{2-4x}}$, allowing us to combine the original and transformed integrals. Summing these expressions eliminates the exponential term and yields $2I = \int_{0}^{1} \frac{dx}{5+2x-2x^2}$.

Evaluating the resulting integral requires completing the square in the quadratic, which gives $5+2x-2x^2 = 2\left( \frac{11}{4} - \left(x - \frac{1}{2}\right)^2 \right)$. The expression becomes $2I = \frac{1}{2} \int_{0}^{1} \frac{dx}{\left(\frac{\sqrt{11}}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2}$. Using the standard integral form $\int \frac{dx}{a^{2-}u^2} = \frac{1}{2a} \ln \left| \frac{a+u}{a-u} \right|$ with $a=\frac{\sqrt{11}}{2}$, we find that $2I = \frac{1}{2} \cdot \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right)^2 = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right)$.

Simplifying the argument of the natural logarithm gives $\frac{\sqrt{11}+1}{\sqrt{11}-1} = \frac{(\sqrt{11}+1)^2}{11-1} = \frac{(\sqrt{11}+1)^2}{10}$, so $2I = \frac{1}{\sqrt{11}} \ln \left( \frac{(\sqrt{11}+1)^2}{10} \right) = \frac{2}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$. Dividing by 2, we determine $I = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$, which identifies the constants as $\alpha = \sqrt{11}$ and $\beta = \sqrt{10}$. Calculating the final value, $\alpha^4 - \beta^4 = (\sqrt{11})^4 - (\sqrt{10})^4 = 121 - 100 = 21$.

The Leibniz integral rule provides the necessary framework for differentiating an integral where the upper limit is a function of the independent variable, stating that the derivative is the integrand evaluated at that limit multiplied by the derivative of the limit itself. Applying this to the given equation, the derivative of the left side with respect to $t$ is $(f(t^2) + (t^2)^2) \cdot (2t)$, while differentiating the right side gives $4t^2$. Equating these results yields $(f(t^2) + t^4) \cdot 2t = 4t^2$.

Simplifying this equation by dividing both sides by $2t$ leads to the expression $f(t^2) = 2t - t^4$. To determine the value of the function at $\frac{\pi^2}{4}$, setting $t^2 = \frac{\pi^2}{4}$ implies $t = \frac{\pi}{2}$. Substituting this value into the derived formula gives $2(\frac{\pi}{2}) - (\frac{\pi}{2})^4$, which simplifies to $\pi - \frac{\pi^4}{16}$. Factoring out $\pi$ results in $\pi \left(1 - \frac{\pi^3}{16}\right)$.

The primary challenge of this integral is the combination of the exponential and logarithmic functions, which is best addressed by the substitution $t = e^x$. This choice simplifies the differential $e^x dx$ into $dt$ and updates the integration boundaries to range from 1/2 to 2. The integrand then takes the form $\ln(t+\sqrt{1+t^2})$, which is a classic candidate for integration by parts.

Setting up the definite integral as $\int_{1/2}^2 \ln(t+\sqrt{1+t^2}) dt$, one assigns $u = \ln(t+\sqrt{1+t^2})$ and $dv = dt$. Differentiating $u$ gives $du = \frac{1}{\sqrt{1+t^2}} dt$. Evaluating the product $uv$ at the limits produces $2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln\left(\frac{1+\sqrt{5}}{2}\right)$. The remaining integral, $\int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt$, is computed by recognizing the derivative of $\sqrt{1+t^2}$, which yields the value $\frac{\sqrt{5}}{2}$.

Subtracting this integral result from the boundary expression produces $2 \ln(2+\sqrt{5}) - \frac{1}{2} \ln\left(\frac{1+\sqrt{5}}{2}\right) - \frac{\sqrt{5}}{2}$. To align this with the options, the logarithmic terms are combined using properties of logarithms, specifically converting $2 \ln(2+\sqrt{5})$ into $\ln((2+\sqrt{5})^2)$ and simplifying the square root structure. This algebra reveals the final result as $\ln\left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2}$.

Repeatedly composing the function $f(x)=\frac{x}{(1+ x^{n})^{1/n}}$ reveals a predictable arithmetic progression in the denominator, allowing us to generalize the $n$-th composition as $f^n(x)=\frac{x}{(1+n x^n)^{1/n}}$. Multiplying this by $x^{n-2}$ inside the integral $\int_{0}^{1} x^{n-2} f^n(x) \, dx$ simplifies the integrand to $\frac{x^{n-1}}{(1+n x^n)^{1/n}}$. By applying the substitution $t = 1+n x^n$, where the differential component becomes $x^{n-1} \, dx = \frac{dt}{n^2}$, the integral transforms into $\frac{1}{n^2} \int_{1}^{1+n} t^{-1/n} \, dt$. Computing this definite integral yields $\frac{1}{n(n-1)} \left( (1+n)^{\frac{n-1}{n}} - 1 \right)$. As $n \to \infty$, the numerator term $(1+n)^{\frac{n-1}{n}}$ behaves linearly with respect to $n$, while the denominator $n(n-1)$ grows quadratically, forcing the overall expression to approach 0.

The problem centers on identifying the periodic nature of the integrand. The term $[\sin(\pi x)]$ repeats every 2 units, and the term $e^{[\cos(2\pi x)]}$ repeats every 1 unit, so the entire expression $\left([\sin(\pi x)]+e^{[\cos(2\pi x)]}\right)$ is periodic with a period of 2. Because the integral spans from -3 to 101, a total range of 104 units, it encompasses exactly 52 full periods. We can therefore simplify the total integral $I$ as follows:

$I = 52 \int_{0}^{2} \left([\sin(\pi x)]+e^{[\cos(2\pi x)]}\right) dx$

To evaluate this, we break the integral into two components. For $[\sin(\pi x)]$, the sine function is non-negative on $[0, 1]$, making the floor 0, and it is negative on $[1, 2]$, making the floor -1. This yields an integral value of $-1$. For the term $e^{[\cos(2\pi x)]}$, the floor function $[\cos(2\pi x)]$ is 0 whenever $\cos(2\pi x) \ge 0$ and -1 whenever $\cos(2\pi x) < 0$. Within one period $[0, 1]$, the cosine function is non-negative on $[0, \frac{1}{4}]$ and $[\frac{3}{4}, 1]$, which covers a combined length of $\frac{1}{2}$ where the integrand simplifies to $e^0 = 1$. The interval $(\frac{1}{4}, \frac{3}{4})$ has a length of $\frac{1}{2}$, where the integrand is $e^{-1}$. Summing these segments over the two-unit range results in $2 \times (\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot e^{-1}) = 1 + \frac{1}{e}$. Adding the two components together, $52 \times (-1 + 1 + \frac{1}{e})$ simplifies to $\frac{52}{e}$.

The expression functions as a Riemann sum, which represents the definition of a definite integral as the number of sub-intervals approaches infinity. As $n \to \infty$, the term $\frac{1}{2^n}$ corresponds to the width of each partition $dx$, and the sum $\sum_{r=1}^{2^n-1} \frac{1}{\sqrt{1-\frac{r}{2^n}}}$ effectively forms the area under the curve $f(x) = \frac{1}{\sqrt{1-x}}$ over the interval from 0 to 1. By recognizing the variable substitution $x = \frac{r}{2^n}$, the limit transforms directly into the definite integral $\int_0^1 \frac{1}{\sqrt{1-x}} \, dx$. Applying the symmetry property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ allows the expression to be rewritten as $\int_0^1 \frac{1}{\sqrt{x}} \, dx$, which evaluates to $[2\sqrt{x}]_0^1$ and results in 2.

Evaluating the integral $I = \int_{\pi/4}^{\pi/3} \left( \frac{8 \sin x - \sin 2x}{x} \right) dx$ relies on understanding the monotonicity of the integrand within the defined limits. Because the term $\frac{\sin x}{x}$ decreases strictly on the interval $(\pi/4, \pi/3)$, the integral is bounded by comparing it against the values of the function at its endpoints, which establishes a clear lower bound of $I > \sqrt{3} - \frac{\pi}{6}$. Given this constraint and the behavior of the trigonometric components, the total area under the curve is bounded such that $\frac{5\pi}{12} < I < \frac{\sqrt{2}}{3} \pi$.

The integration of a greatest integer function involving $1/x$ is simplified by using the substitution $t = 1/x$. This change of variables adjusts the differential $dx$ to $-dt/t^2$ and shifts the integration limits from $0$ to $1$ into $\infty$ to $1$. Consequently, the integral transforms into $\int_1^\infty \frac{1}{t^2 7^{[t]}} dt$. Since the greatest integer function $[t]$ maintains a constant value between consecutive integers, the integral can be represented as an infinite sum of integrals over the intervals $[n, n+1]$ for $n=1, 2, 3, \dots$, leading to the expression $\sum_{n=1}^\infty \int_n^{n+1} \frac{1}{7^n t^2} dt$.

Evaluating the integral within the summation gives $\sum_{n=1}^\infty \frac{1}{7^n} \left( \frac{1}{n} - \frac{1}{n+1} \right)$, which further expands into two distinct series: $\sum_{n=1}^\infty \frac{1}{n 7^n} - \sum_{n=1}^\infty \frac{1}{(n+1) 7^n}$. By applying the standard Taylor series expansion for the natural logarithm, $-\ln(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$ with $z=1/7$, the first series becomes $-\ln(1 - 1/7) = -\ln(6/7)$. The second series, after adjusting the index and factor, can be expressed as $7 \sum_{n=1}^\infty \frac{(1/7)^{n+1}}{n+1} = 7 \left( -\ln(1 - 1/7) - 1/7 \right)$, which simplifies to $-7 \ln(6/7) - 1$. Subtracting this second series from the first leads to $-\ln(6/7) - (-7 \ln(6/7) - 1)$, resulting in the final value of $1 + 6 \ln(6/7)$.

Applying the Leibniz integral rule to differentiate the given equation $f(x) + \int_{0}^{x}(x-t)f'(t)dt = (e^{2x}+e^{-2x})\cos 2x + \frac{2}{a}x$ with respect to $x$ yields f'(x) + \int_{0}^{x} f'(t) dt = \frac{d}{dx} \[(e^{2x}+e^{-2x})\cos 2x]$ + \frac{2}{a}$. Since the integral term$\int_{0}^{x} f'(t) dt$simplifies to$f(x) - f(0)$, the equation becomes$f'(x) + f(x) - f(0) = \frac{d}{dx} $[(e^{2x}+e^{-2x})\cos 2x]$ + \frac{2}{a}$. Evaluating the original equation at$x=0$provides$f(0) + 0 = (e^{0}+e^{0})\cos 0 + 0$, which results in$f(0) = 2$. Substituting$f(0)=2$and the given condition$f'(0)=4$into the differentiated equation at$x=0$gives$4 + 2 - 2 = \left. \frac{d}{dx} $[(e^{2x}+e^{-2x})\cos 2x]$ \right|_{x=0} + \frac{2}{a}$. Defining$H(x) = (e^{2x}+e^{-2x})\cos 2x$, the derivative is$H'(x) = (2e^{2x}-2e^{-2x})\cos 2x - 2(e^{2x}+e^{-2x})\sin 2x$. At$x=0$,$H'(0) = (2-2)(1) - 2(1+1)(0) = 0$. Consequently,$4 = 0 + \frac{2}{a}$, which implies$a = \frac{1}{2}$. Finally, calculating$(2a+1)^{5}a^{2}$using$a=\frac{1}{2}$results in$(2(\frac{1}{2})+1)^{5}(\frac{1}{2})^{2} = (2)^{5} \cdot \frac{1}{4} = \frac{32}{4} = 8$.

Because the expression inside the cosine involves the greatest integer function $\lfloor \frac{x}{2} \rfloor$, we must partition the integration interval $[0, 5]$ into sub-intervals where the floor function remains constant. Since $\lfloor \frac{x}{2} \rfloor$ changes value at every even integer, the points $x=2$ and $x=4$ create three distinct regions: $[0, 2)$, $[2, 4)$, and $[4, 5]$.

Within these regions, the floor function behaves predictably: it equals $0$ on $[0, 2)$, $1$ on $[2, 4)$, and $2$ on $[4, 5]$. Substituting these values transforms the integral into the sum $\int_0^2 \cos(\pi x) \, dx + \int_2^4 \cos(\pi(x-1)) \, dx + \int_4^5 \cos(\pi(x-2)) \, dx$. Using trigonometric identities, specifically $\cos(\theta - \pi) = -\cos(\theta)$ and $\cos(\theta - 2\pi) = \cos(\theta)$, this simplifies to $\int_0^2 \cos(\pi x) \, dx - \int_2^4 \cos(\pi x) \, dx + \int_4^5 \cos(\pi x) \, dx$. Integrating $\cos(\pi x)$ yields $\frac{\sin(\pi x)}{\pi}$, and evaluating this antiderivative at the boundaries $0, 2, 4, 5$ results in zero for each segment, leading to a total result of $0$.

To determine the relationship between the terms, we begin by evaluating the difference $b_n - b_{n-1}$, which is represented by the integral $\int_{0}^{\pi/2} \frac{\cos^2 nx - \cos^2 (n-1)x}{\sin x} \, dx$. By applying the trigonometric identity $\cos^2 A - \cos^2 B = -\sin(A-B)\sin(A+B)$, the expression in the numerator simplifies to $-\sin(nx - (n-1)x)\sin(nx + (n-1)x)$, which is equivalent to $-\sin x \sin(2n-1)x$. Dividing this by the denominator $\sin x$ reduces the integrand to simply $-\sin(2n-1)x$.