Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The integral equation $f(x)=e^x+\int_{0}^{1} (y+x e^x) f(y) d y$ is solved by decomposing the integral into constant parameters $A=\int_{0}^{1} y f(y) d y$ and $B=\int_{0}^{1} f(y) d y$, which transforms the expression into $f(x)=e^x(1+x B)+A$. Substituting $f(y)=e^y(1+y B)+A$ into the expression for $B$ yields $B=\int_{0}^{1} (e^y+A+y B e^y) d y = (e-1)+A+B$, where the equality \int_{0}^{1} y e^y d y = \[y e^y - e^y]_0^1 = 1$is applied. Subtracting$B$from both sides results \in$0 = e - 1 + A$, confirming that$A = 1 - e$. Given that$f(0)=e^0(1+0 \cdot B)+A = 1+A$, substituting$A=1-e$leads to$f(0)=2-e$. Consequently, the value of$e+f(0)$is$e+(2-e)=2$.

The integrand $\sin^4 x + \cos^4 x$ can be simplified using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$, which reduces to $1 - 2 \sin^2 x \cos^2 x$. Since $2 \sin x \cos x = \sin 2x$, this expression further simplifies to $1 - \frac{1}{2} \sin^2 2x$. Integrating this over the interval from $0$ to $20\pi$ gives $\int_{0}^{20\pi} 1 \, dx - \frac{1}{2} \int_{0}^{20\pi} \sin^2 2x \, dx$. The first part yields $20\pi$. For the second part, the function $\sin^2 2x$ has a period of $\frac{\pi}{2}$, meaning there are $40$ cycles within the interval of $20\pi$. Because the integral of $\sin^2 2x$ over one period $\frac{\pi}{2}$ is $\frac{\pi}{4}$, the total integral over $40$ periods is $40 \times \frac{\pi}{4} = 10\pi$. Multiplying by the constant $\frac{1}{2}$ results in $5\pi$, and subtracting this from $20\pi$ leaves $15\pi$.

To find the polynomial $f(x)$, substitute $t = x^{2+}1$, which implies $x^2 = t-1$. Substituting this into the given equation $f(x^{2+}1) = x^{4+}5x^{2+}2$ results in $f(t) = (t-1)^2 + 5(t-1) + 2$. Expanding this expression gives $f(t) = t^2 - 2t + 1 + 5t - 5 + 2$, which simplifies to $f(t) = t^2 + 3t - 2$, so $f(x) = x^2 + 3x - 2$.

The definite integral is evaluated as $\int_{0}^{3} (x^{2+}3x-2) \, dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_{0}^{3}$. Substituting the upper limit yields $\left( \frac{27}{3} + \frac{3(9)}{2} - 2(3) \right) - 0 = 9 + \frac{27}{2} - 6 = 3 + \frac{27}{2} = \frac{6+27}{2} = \frac{33}{2}$.

The degree of the polynomial $n$ is determined by comparing the powers in the identity $f(x) = f'(x)f''(x)$, which requires $n = (n-1) + (n-2)$, simplifying to $n=3$. Considering $f(0)=0$, we can express the cubic polynomial as $f(x) = ax^3 + bx^2 + cx$. Computing the derivatives $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$, we equate their product to $f(x)$ by expanding $(3ax^2 + 2bx + c)(6ax + 2b) = 18a^2x^3 + 18abx^2 + (6ac + 4b^2)x + 2bc$. Comparing this expansion with $f(x) = ax^3 + bx^2 + cx$ implies $18a^2 = a$, leading to $a = 1/18$ because $a \neq 0$. The constant term $2bc = 0$ implies either $b=0$ or $c=0$, and subsequent analysis of the coefficients for $x^2$ and $x$ confirms that both must be zero, which reduces the polynomial to $f(x) = \frac{x^3}{18}$.

With the polynomial defined, the derivative $f'(2)$ is $\frac{2^2}{6} = \frac{2}{3}$, and the second derivative $f''(2)$ is $\frac{2}{3}$. The definite integral is computed as $\int_{0}^{2} \frac{x^3}{18} dx = \left[ \frac{x^4}{72} \right]_{0}^{2} = \frac{16}{72} = \frac{2}{9}$. Summing these values gives $\frac{2}{3} + \frac{2}{3} + \frac{2}{9} = \frac{14}{9}$. Multiplying this result by 36 yields $36 \times \frac{14}{9} = 56$.

Since the integrand $f(x) = | \sin x | + [x \sin x]$ is an even function, the integral from $-2$ to $2$ is equal to $2 \int_{0}^{2} (\sin x + [x \sin x]) \, dx$. Within the interval $[0, 2]$, $\sin x$ is non-negative, so the absolute value can be removed. The function $x \sin x$ is strictly increasing in this interval, and we define $\alpha$ as the unique value in $(0, 2)$ where $x \sin x = 1$. Consequently, the greatest integer $[x \sin x]$ equals $0$ for $x \in [0, \alpha)$ and $1$ for $x \in [\alpha, 2]$.

Evaluating the integral yields $2 \left( \int_{0}^{2} \sin x \, dx + \int_{\alpha}^{2} 1 \, dx \right)$. Calculating the definite integrals gives $2 \left( [-\cos x]_{0}^{2} + [x]_{\alpha}^{2} \right)$, which simplifies to $2 ((1 - \cos 2) + (2 - \alpha))$. Distributing the factor of $2$ results in $2(3 - \cos 2) - 2\alpha$. Comparing this expression to the given form $2(3 - \cos 2) + \beta$, it is evident that $\beta = -2\alpha$. Given the condition $\alpha \sin \alpha = 1$, substituting $\alpha = -\beta/2$ leads to $(-\beta/2) \sin(-\beta/2) = 1$, which simplifies to $\beta \sin \left( \frac{\beta}{2} \right) = 2$.

The substitution $x = 2\tan\theta$ is ideal here because the denominator $x^{2+}4$ simplifies to $4\sec^2\theta$, which cleanly cancels out the differential $dx = 2\sec^2\theta d\theta$. As $x$ moves from $0$ to $\infty$, the corresponding limits for $\theta$ are $0$ and $\frac{\pi}{2}$. Applying this transformation, the integral becomes $\int_{0}^{\frac{\pi}{2}} \frac{\log_e(2\tan\theta)}{4\sec^2\theta} \cdot 2\sec^2\theta d\theta$, simplifying to $\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log_e(2\tan\theta) d\theta$.

Breaking this into logarithmic parts, we get $\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log_e(2) d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log_e(\tan\theta) d\theta$. The constant term evaluates directly to $\frac{\pi \log_e(2)}{4}$, while the term $\int_{0}^{\frac{\pi}{2}} \log_e(\tan\theta) d\theta$ evaluates to zero because substituting $\theta = \frac{\pi}{2} - u$ shows the integral is its own negative. This yields the result $\frac{\pi \log_e(2)}{4}$.

The integral $\int_{0}^{r} x |\sin \pi x| \, dx$ can be evaluated by decomposing the interval into sub-intervals of length unity, expressed as $\sum_{k=0}^{r-1} \int_{k}^{k+1} x |\sin \pi x| \, dx$. Within each unit interval $[k, k+1]$, the substitution $x = u+k$ yields $\int_{0}^{1} (u+k) |\sin \pi(u+k)| \, du = \int_{0}^{1} (u+k) \sin \pi u \, du$ because $\sin \pi(u+k) = (-1)^k \sin \pi u$ and the absolute value evaluates to $\sin \pi u$ for $u \in [0, 1]$. Calculating this definite integral results in $\int_{0}^{1} u \sin \pi u \, du + k \int_{0}^{1} \sin \pi u \, du = \frac{1}{\pi} + \frac{2k}{\pi} = \frac{2k+1}{\pi}$.

Summing these individual integrals from $k=0$ to $r-1$ gives the total integral $I_r = \sum_{k=0}^{r-1} \frac{2k+1}{\pi} = \frac{r^2}{\pi}$, based on the property that the sum of the first $r$ odd integers is $r^2$. Substituting this result back into the original expression gives $\sum_{r=1}^{20} \sqrt{\pi \cdot \frac{r^2}{\pi}} = \sum_{r=1}^{20} r$. Applying the formula for the sum of the first $n$ integers, $\sum_{r=1}^{20} r = \frac{20(21)}{2} = 210$.

The property of definite integrals states that for an odd function $f(x)$ over a symmetric interval $[-a, a]$, the integral evaluates to zero. Conversely, for an even function, the integral over a symmetric interval $[-a, a]$ is equal to twice the integral over the interval $[0, a]$.

Splitting the integrand yields $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\pi}{1-\sin(|x|+\frac{\pi}{6})} dx + \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{4 x^{11}}{1-\sin(|x|+\frac{\pi}{6})} dx$. Since the second integrand contains an odd function $x^{11}$ in the numerator and an even function in the denominator, the entire expression is an odd function, meaning this part of the integral evaluates to zero. The first part is an even function, so the integral simplifies to $2\pi \int_{0}^{\frac{\pi}{6}} \frac{dx}{1-\sin(x+\frac{\pi}{6})}$.

Multiplying the numerator and denominator by $1+\sin(x+\frac{\pi}{6})$ results in $2\pi \int_{0}^{\frac{\pi}{6}} \frac{1+\sin(x+\frac{\pi}{6})}{\cos^2(x+\frac{\pi}{6})} dx = 2\pi \int_{0}^{\frac{\pi}{6}} (\sec^2(x+\frac{\pi}{6}) + \sec(x+\frac{\pi}{6})\tan(x+\frac{\pi}{6})) dx$. Evaluating the antiderivative gives $2\pi [\tan(x+\frac{\pi}{6}) + \sec(x+\frac{\pi}{6})]_{0}^{\frac{\pi}{6}}$, which is $2\pi [(\tan\frac{\pi}{3} + \sec\frac{\pi}{3}) - (\tan\frac{\pi}{6} + \sec\frac{\pi}{6})]$. Substituting the trigonometric values $2\pi [(\sqrt{3} + 2) - (\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})]$ simplifies to $2\pi [\sqrt{3} + 2 - \sqrt{3}]$, which equals $4\pi$.

The integrand $\cot\left(x - \frac{\pi}{3}\right) \cot\left(x + \frac{\pi}{3}\right) + 1$ can be simplified by applying the trigonometric identity for the cotangent of a difference, specifically $\cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$. By setting $A = x + \frac{\pi}{3}$ and $B = x - \frac{\pi}{3}$, the difference $A - B$ becomes $\frac{2\pi}{3}$, and since $\cot\left(\frac{2\pi}{3}\right) = -\frac{1}{\sqrt{3}}$, the integrand transforms into $-\frac{1}{\sqrt{3}} \left[ \cot\left(x - \frac{\pi}{3}\right) - \cot\left(x + \frac{\pi}{3}\right) \right]$.

Integrating this expression term by term results in $-\frac{1}{\sqrt{3}} \left[ \ln \left| \sin\left(x - \frac{\pi}{3}\right) \right| - \ln \left| \sin\left(x + \frac{\pi}{3}\right) \right| \right]$, which evaluates to $-\frac{1}{\sqrt{3}} \ln \left| \frac{\sin(x - \pi/3)}{\sin(x + \pi/3)} \right|$ over the interval from $\frac{\pi}{6}$ to $\frac{\pi}{4}$. Evaluating at the upper limit of $x = \frac{\pi}{4}$ yields $\left| \frac{\sin(-\pi/12)}{\sin(7\pi/12)} \right| = \tan(15^\circ) = 2 - \sqrt{3}$, and at the lower limit of $x = \frac{\pi}{6}$ yields $\left| \frac{\sin(-\pi/6)}{\sin(\pi/2)} \right| = \frac{1/2}{1} = \frac{1}{2}$. The definite integral is therefore $-\frac{1}{\sqrt{3}} \left( \ln(2 - \sqrt{3}) - \ln(1/2) \right)$, which simplifies to $-\frac{1}{\sqrt{3}} \ln(4 - 2\sqrt{3})$.

Recognizing that $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$, the expression becomes $-\frac{2}{\sqrt{3}} \ln(\sqrt{3} - 1)$. Comparing this to the form $\alpha \log_e(\sqrt{3} - 1)$, it is clear that $\alpha = -\frac{2}{\sqrt{3}}$. Squaring this coefficient and multiplying by $9$ gives $9 \left( \frac{4}{3} \right)$, resulting in a final value of $12$.

The integral equation involves a scaling variable $\lambda$ inside the function $f(\lambda x)$, suggesting a transformation to relate the antiderivative directly to the function. By substituting $u = \lambda x$ and $d\lambda = \frac{1}{x}du$, the given expression transforms into $\frac{1}{x} \int_{0}^{x} f(u) du = a f(x)$. Applying the fundamental theorem of calculus by differentiating both sides with respect to $x$ yields the following differential relationship:
$\frac{f'(x)}{f(x)} = \frac{1-a}{ax}$

Integrating this expression gives $\ln|f(x)| = \frac{1-a}{a} \ln x + C$, and since $f(1)=1$, the constant $C$ is zero, leading to the general form $f(x) = x^{(1-a)/a}$. Using the condition $f(16) = 1/8$, we find $16^{(1-a)/a} = 2^{-3}$, which implies $(2^4)^{(1-a)/a} = 2^{-3}$, resulting in the equation $4(1-a)/a = -3$ and consequently $a=4$. With this value, the function simplifies to $f(x) = x^{-3/4}$, which has a derivative $f'(x) = -\frac{3}{4} x^{-7/4}$. Substituting $x = 1/16$ into the derivative gives $f'(1/16) = -\frac{3}{4} (2^{-4})^{-7/4} = -\frac{3}{4} (128) = -96$, so the final expression $16 - f'(1/16)$ evaluates to $16 - (-96)$, which is 112.

Solving this integral relies on the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, which helps eliminate the linear term $x$ from the numerator. Denoting the integral as $I$, we rewrite it by replacing $x$ with $\pi - x$ in the integrand to get $I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 x + \sin^2 x} dx$, noting that the denominator remains unchanged due to the properties of trigonometric functions.

Adding the two expressions for $I$ yields
$2I = \int_{0}^{\pi} \frac{8\pi}{4 \cos^2 x + \sin^2 x} dx$
which simplifies to $I = 4\pi \int_{0}^{\pi} \frac{1}{4 \cos^2 x + \sin^2 x} dx$. Utilizing the symmetry of the integrand about $\frac{\pi}{2}$, we can compute the integral from $0$ to $\frac{\pi}{2}$ and multiply by two, giving $I = 8\pi \int_{0}^{\pi/2} \frac{\sec^2 x}{4 + \tan^2 x} dx$. Substituting $t = \tan x$ with $dt = \sec^2 x dx$ changes the integration limits to $0$ and $\infty$, resulting in
$I = 8\pi \int_{0}^{\infty } \frac{dt}{4 + t^2}$
Evaluating this integral using the standard arctangent formula $8\pi \cdot \left[ \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) \right]_{0}^{\infty }$ yields $8\pi \cdot \frac{1}{2} \cdot \frac{\pi}{2}$, which results in $2\pi^2$.

The limit $\lim_{t\to 0} \left(\int_0^1 (3x+5)^t \, dx\right)^{1/t}$ presents an indeterminate form of $1^\infty$. Applying the standard result for such limits, the expression evaluates to $e^{\lim_{t\to 0} \frac{1}{t}(\int_0^1 (3x+5)^t \, dx - 1)}$, which simplifies to $e^{\int_0^1 \ln(3x+5) \, dx}$ using L'Hôpital's rule. Evaluating the integral with the substitution $u = 3x+5$ yields $\frac{1}{3}\int_5^8 \ln u \, du$, which equates to $\frac{8 \ln 8 - 5 \ln 5 - 3}{3}$.

This exponent can be expanded as $\ln(8^{8/3}) - \ln(5^{5/3}) - 1$, and raising $e$ to this value produces $\frac{8^{8/3}}{5^{5/3}e}$. Through algebraic manipulation, this fraction is rewritten as $\frac{64}{5e}\left(\frac{8}{5}\right)^{2/3}$. Matching this form to the provided expression $\frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$ yields $\alpha = 64$.

The definite integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} \, dx$ can be efficiently evaluated using the symmetry property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx$. Since the sum of the integrand at $x$ and $-x$ simplifies as $\frac{96 x^2 \cos^2 x}{1+e^x} + \frac{96 x^2 \cos^2 x}{1+e^{-x}} = 96 x^2 \cos^2 x$, the integral reduces to $96 \int_{0}^{\frac{\pi}{2}} x^2 \cos^2 x \, dx$.

Using the trigonometric identity $\cos^2 x = \frac{1+\cos 2x}{2}$, the expression becomes $48 \int_{0}^{\frac{\pi}{2}} (x^2 + x^2 \cos 2x) \, dx$. Integrating the first term, $48 \int_{0}^{\frac{\pi}{2}} x^2 \, dx$, yields $48 \left[ \frac{x^3}{3} \right]_{0}^{\frac{\pi}{2}} = 2\pi^3$. The second part, $48 \int_{0}^{\frac{\pi}{2}} x^2 \cos 2x \, dx$, is solved through two applications of integration by parts, which results in $-12\pi$. Combining these values gives $2\pi^3 - 12\pi$, which can be written in the form $\pi (2 \pi^2 - 12)$. By comparing this to the given expression $\pi (\alpha \pi^2 + \beta)$, we identify $\alpha = 2$ and $\beta = -12$. Consequently, $(\alpha + \beta)^2 = (2 - 12)^2$, which equals 100.

To evaluate the integral of the greatest integer function, we must determine the points at which the expression $\frac{1}{e^{x-1}}$, or $e^{1-x}$, passes through integer values. As $x$ increases from $0$ to $e^3$, the function $e^{1-x}$ decreases from $e$ to $e^{1-e^3}$. We identify the intervals where the floor of this expression remains constant: when $x$ is between $0$ and $1 - \ln 2$, $e^{1-x}$ lies between $2$ and $e$, so the greatest integer is $2$. When $x$ is between $1 - \ln 2$ and $1$, the expression lies between $1$ and $2$, making the greatest integer $1$. Finally, for $x$ greater than $1$ up to $e^3$, the value is less than $1$ but greater than $0$, so the greatest integer is $0$.

We calculate the integral by summing the areas of these specific regions, defined as $\int_{0}^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^{1} 1 \, dx + \int_{1}^{e^3} 0 \, dx$. Evaluating the first integral yields $2(1 - \ln 2)$, and the second integral yields $1 - (1 - \ln 2)$, which simplifies to $\ln 2$. Combining these results gives $2 - 2\ln 2 + \ln 2$, resulting in $2 - \ln 2$. Matching this to the expression $\alpha - \log_{e} 2$, we see that $\alpha$ must be $2$, and consequently, $\alpha^{3}$ equals $8$.

The range of a continuous function on a closed interval is defined by its values at the endpoints and at any points where the derivative vanishes. Integrating the provided expression gives $f(x) = \int_0^x (t^3 - 9t^2 + 20t) \, dt = \frac{x^4}{4} - 3x^3 + 10x^2$, and setting the derivative $f'(x) = x(x-4)(x-5)$ to zero identifies $x=4$ as the critical point within the interval $[1, 5]$.

Evaluating the function at these critical and endpoint locations provides the necessary values: $f(1) = \frac{29}{4}$, $f(4) = 32$, and $f(5) = \frac{125}{4}$. Comparing these outcomes, the minimum value is $\alpha = \frac{29}{4}$ and the maximum value is $\beta = 32$. Finally, the calculation $4(\alpha + \beta)$ becomes $4\left(\frac{29}{4} + 32\right) = 29 + 128 = 157$.

The integral $\int_{0}^{\frac{\pi}{4}} \sin \left|4x - \frac{\pi}{12}\right| dx$ requires splitting the domain at $x = \frac{\pi}{48}$ because the expression inside the absolute value changes sign at this point. In the interval $\left[0, \frac{\pi}{48}\right]$, the integrand is $-\sin \left(4x - \frac{\pi}{12}\right)$, while for $\left[\frac{\pi}{48}, \frac{\pi}{4}\right]$, it is $\sin \left(4x - \frac{\pi}{12}\right)$. Performing these integrations leads to a total value of $0.5$ for this part, and when scaled by the external factor of $24$, the result is $12$. The greatest integer function $[2 \sin x]$ remains $0$ as long as $2 \sin x < 1$, which is equivalent to $\sin x < 0.5$, occurring for $x < \frac{\pi}{6}$. For $x \ge \frac{\pi}{6}$, the value of $2 \sin x$ reaches at least $1$ but stays below $2$, so $[2 \sin x] = 1$. Thus, the integral $\int_{0}^{\frac{\pi}{4}} [2 \sin x] dx$ simplifies to $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 dx = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$. Multiplying this result by $24$ yields $2\pi$. Adding the two calculated components, $12 + 2\pi$, directly identifies $\alpha$ as $12$.

The integrand can be simplified by distributing the denominator across the numerator, resulting in the sum $\frac{e^x}{e^x+e^{-x}} + \sqrt{|x|-x}$. The first integral $\int_{-1}^{1} \frac{e^x}{e^x+e^{-x}} dx$ evaluates to 1 by applying the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, which yields $2I_1 = \int_{-1}^{1} \frac{e^x+e^{-x}}{e^x+e^{-x}} dx = \int_{-1}^{1} 1 dx = 2$. For the second integral $\int_{-1}^{1} \sqrt{|x|-x} dx$, the integrand becomes zero on $[0, 1]$ because $|x|=x$, and simplifies to $\sqrt{-2x}$ on $[-1, 0]$ because $|x|=-x$. Evaluating the non-zero portion gives $\int_{-1}^{0} \sqrt{-2x} dx$, which upon substituting $t = -x$ becomes $\sqrt{2} \int_{0}^{1} \sqrt{t} dt = \sqrt{2} \left[ \frac{2}{3} t^{3/2} \right]_0^1 = \frac{2\sqrt{2}}{3}$. Combining these two results, the total value of the definite integral is $1 + \frac{2\sqrt{2}}{3}$.

The defining characteristic of this integral is the application of the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$, which effectively simplifies the expression by exploiting the symmetry of the limits $[0, \pi]$. Given the integral $I = \int_{0}^{\pi} \frac{(x+3) \sin x}{1+3 \cos^2 x} \, dx$, replacing $x$ with $\pi - x$ creates a second expression, $I = \int_{0}^{\pi} \frac{(\pi - x + 3) \sin(\pi - x)}{1+3 \cos^2(\pi - x)} \, dx$. Because $\sin(\pi - x)$ equals $\sin x$ and $\cos(\pi - x)$ equals $-\cos x$, the trigonometric components remain invariant, resulting in $I = \int_{0}^{\pi} \frac{(\pi - x + 3) \sin x}{1+3 \cos^2 x} \, dx$.

Summing the original integral and this transformed version leads to $2I = \int_{0}^{\pi} \frac{(\pi + 6) \sin x}{1+3 \cos^2 x} \, dx$, which neatly eliminates the variable $x$ from the numerator. This constant factor $(\pi + 6)$ can be moved outside the integral, leaving us to evaluate $\int_{0}^{\pi} \frac{\sin x}{1+3 \cos^2 x} \, dx$. By using the substitution $t = \cos x$, where $dt = -\sin x \, dx$, the limits change from $0$ to $\pi$ into $1$ to $-1$. Applying these bounds, the expression becomes $(\pi+6) \int_{-1}^{1} \frac{dt}{1+3t^2}$.