Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The expansion of the squared term $(|\sin x|+|\cos x|)^2$ results in $\sin^2 x + \cos^2 x + 2|\sin x \cos x|$, which simplifies to $1 + |\sin 2x|$ by applying the identity $\sin^2 x + \cos^2 x = 1$ and the double-angle property. Because the function $1 + |\sin 2x|$ is periodic with a period of $\frac{\pi}{2}$, the integral over the range from $0$ to $20\pi$ effectively spans 40 such periods. This allows the integral to be expressed as $40 \int_{0}^{\frac{\pi}{2}} (1 + \sin 2x) dx$. Evaluating the antiderivative $x - \frac{\cos 2x}{2}$ at the upper bound of $\frac{\pi}{2}$ yields $\frac{\pi}{2} - \frac{\cos \pi}{2} = \frac{\pi}{2} + \frac{1}{2}$, and evaluating at the lower bound of $0$ yields $0 - \frac{\cos 0}{2} = -\frac{1}{2}$. The difference between these bounds results in $\frac{\pi}{2} + \frac{1}{2} - (-\frac{1}{2}) = \frac{\pi}{2} + 1$. Multiplying this value by the factor of 40 yields $20\pi + 40$, which simplifies to $20(\pi + 2)$.

The integral relies on the symmetry property of definite integrals, specifically $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx$ when applied to an even function. Given the integrand $f(x) = \frac{1}{(1+e^x)(\sin^6 x + \cos^6 x)}$, we can pair it with its reflection across the y-axis, $f(-x) = \frac{1}{(1+e^{-x})(\sin^6 x + \cos^6 x)}$, noting that the trigonometric denominator is an even function. Adding these terms, the exponential expressions combine as $\frac{1}{1+e^x} + \frac{1}{1+e^{-x}} = \frac{1}{1+e^x} + \frac{e^x}{e^x+1} = 1$, which simplifies the entire integral to $I = \int_{0}^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$.

To evaluate this remaining form, rewrite the denominator using the identity $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3 \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$, which reduces to $1 - 3 \sin^2 x \cos^2 x$, or $1 - \frac{3}{4} \sin^2 2x$. The integral becomes $\int_{0}^{\pi/2} \frac{4 \, dx}{4 - 3 \sin^2 2x}$. Dividing the numerator and denominator by $\cos^2 2x$ transforms the integrand into $\frac{4 \sec^2 2x}{4 \sec^2 2x - 3 \tan^2 2x}$, and substituting $\sec^2 2x = 1 + \tan^2 2x$ yields $\int_{0}^{\pi/2} \frac{4 \sec^2 2x \, dx}{4 + \tan^2 2x}$. By symmetry around $\pi/4$, this equals $2 \int_{0}^{\pi/4} \frac{4 \sec^2 2x \, dx}{4 + \tan^2 2x}$. Applying the substitution $t = \tan 2x$, where $dt = 2 \sec^2 2x \, dx$, the integral transforms into $2 \int_{0}^{\infty } \frac{2 \, dt}{4 + t^2} = 4 \int_{0}^{\infty } \frac{dt}{4 + t^2}$. Evaluating this gives $4 \left( \frac{1}{2} \tan^{-1} \frac{t}{2} \right)_0^{\infty } = 2 \left( \frac{\pi}{2} \right) = \pi$.

The core concept behind solving integral equations of this form is recognizing that integrals over fixed limits yield constant values rather than variables. By treating these integral terms as unknown constants, the expression simplifies to a linear function $f(x) = Ax - B$. This transformation converts the complex integral equation into a manageable system of simultaneous algebraic equations, where the coefficients $A$ and $B$ are defined by the integral terms themselves.

Applying the integral properties, the original equation expands to $f(x) = x + x \int_0^1 f(t) dt - \int_0^1 t f(t) dt$. Defining the constants as $A = 1 + \int_0^1 f(t) dt$ and $B = \int_0^1 t f(t) dt$, the function is reduced to $f(x) = Ax - B$. Substituting this form back into the integral definitions results in $A = 1 + \int_0^1 (At - B) dt$ and $B = \int_0^1 t(At - B) dt$. Evaluating these integrals yields the equations $A = 2(1 - B)$ and $A = \frac{9}{2}B$. Solving this system for the two variables leads to $A = \frac{18}{13}$ and $B = \frac{4}{13}$.

With these constants determined, the function is $f(x) = \frac{18}{13}x - \frac{4}{13}$. Testing the coordinates $(6, 8)$, we find that for $x=6$, the function value calculates to $f(6) = \frac{18}{13}(6) - \frac{4}{13} = \frac{108 - 4}{13} = 8$. This result confirms that the point satisfies the curve equation.

Integration by parts serves as the primary tool for evaluating integrals of the form $I_n(x) = \int_0^x (t^{2+}5)^{-n} dt$ because it allows for the derivation of a recursive reduction formula. By choosing $u=(t^{2+}5)^{-n}$ and $dv=dt$, the integral expands to $\frac{x}{(x^{2+}5)^n} + 2n \int_0^x t^2(t^{2+}5)^{-n-1} dt$. Splitting the numerator $t^2$ into $(t^{2+}5)-5$ transforms the remaining integral into $2n(I_n(x) - 5I_{n+1}(x))$, resulting in the identity $I_n(x) = \frac{x}{(x^{2+}5)^n} + 2nI_n(x) - 10nI_{n+1}(x)$.

Rearranging this identity demonstrates that $10nI_{n+1}(x) = (2n-1)I_n(x) + x(x^{2+}5)^{-n}$. Utilizing the definition $I_n'(x) = (x^{2+}5)^{-n}$, the relationship simplifies to $10nI_{n+1}(x) - (2n-1)I_n(x) = xI_n'(x)$. Applying this result for $n=5$ yields $50I_6(x) - 9I_5(x) = xI_5'(x)$, which corresponds to the expression $50I_6 - 9I_5 - xI_5'$.

The function $f(x) = \min\{[x-1], [x-2], \dots, [x-10]\}$ simplifies because the sequence of greatest integer terms $[x-1], [x-2], \dots, [x-10]$ is monotonically decreasing, meaning the smallest value is always $[x-10]$. Over the domain $0 \leq x \leq 10$, the expression $[x-10]$ yields non-positive integer values, which implies that the absolute value $|f(x)|$ equals $-f(x)$. Consequently, the expression $\int_{0}^{10} f(x) dx + \int_{0}^{10} (f(x))^2 dx + \int_{0}^{10} |f(x)| dx$ simplifies by canceling the integral of $f(x)$ with the integral of $|f(x)|$, leaving only the integral of $(f(x))^2$.

Evaluating $\int_{0}^{10} [x-10]^2 dx$ requires summing the squares of the integers from $-10$ to $-1$, as the step function $[x-10]$ takes constant values of $-10, -9, \dots, -1$ over each unit interval within $[0, 10]$. Summing these squared values yields $10^2 + 9^2 + \dots + 1^2$, which is the sum of the squares of the first ten integers. Using the formula $\frac{n(n+1)(2n+1)}{6}$ for $n=10$, we arrive at the final value:
$\frac{10 \times 11 \times 21}{6} = 385$

To determine the value of the given integral, we first examine the behavior of $f(x)$ over unit intervals based on the floor function $[x]$. For $0 \leq x < 1$, where $[x]=0$ is even, the function is defined as $f(x) = 1-x$. In the next interval, $1 \leq x < 2$, where $[x]=1$ is odd, the definition changes to $f(x) = x-1$. This alternating pattern continues, which confirms that $f(x)$ is periodic with a period of 2. Because $\cos(\pi x)$ also shares this period of 2, the entire expression inside the integral repeats every 2 units.

Given that the integration bounds extend from $-10$ to $10$, covering a total interval length of 20, we can simplify the integral to 10 times the integral over any single period of length 2, such as $[0, 2]$. Calculating $\int_{0}^{2} f(x) \cos(\pi x) \, dx$ requires splitting it into the sum of $\int_{0}^{1} (1-x) \cos(\pi x) \, dx$ and $\int_{1}^{2} (x-1) \cos(\pi x) \, dx$. Using integration by parts, the first integral evaluates to $\frac{2}{\pi^2}$ and the second also evaluates to $\frac{2}{\pi^2}$, resulting in a total of $\frac{4}{\pi^2}$ for the single period. Multiplying this sum by the factor $\frac{\pi^2}{10}$ and the 10 repetitions across the interval $[-10, 10]$ yields the final result of 4.

The derivative $f^{\prime}(t)$ is obtained directly from the Leibniz rule as $f^{\prime}(t)=e^{t^{3}}\frac{t^{8}}{(t^{6}+2t^{3}+2)^{2}}$, which simplifies at $t=1$ to $f^{\prime}(1)=\frac{e}{25}$. To find $f(1)$, we apply the substitution $z=x^{3}$, which implies $dz=3x^{2}dx$ and transforms the integral into $\frac{1}{3}\int_{0}^{t^{3}} e^{z} \frac{z^{2}}{(z^{2}+2z+2)^{2}} dz$. By expressing the numerator $z^{2}$ as $(z^{2}+2z+2)-(2z+2)$, the integrand splits into the form $e^{z}(g(z)+g^{\prime}(z))$, where $g(z)=\frac{1}{z^{2}+2z+2}$ and $g^{\prime}(z)=-\frac{2z+2}{(z^{2}+2z+2)^{2}}$. Utilizing the integration rule $\int e^{z}(g(z)+g^{\prime}(z))dz = e^{z}g(z)$ produces $f(t)=\frac{1}{3}\left(\frac{e^{t^{3}}}{t^{6}+2t^{3}+2}-\frac{1}{2}\right)$, and evaluating this at $t=1$ gives $f(1)=\frac{e}{15}-\frac{1}{6}$. Substituting these values into the condition $f(1)+f^{\prime}(1)=\alpha e-\frac{1}{6}$ leads to $\frac{e}{15}+\frac{e}{25}=\alpha e$, which simplifies to $\frac{8e}{75}=\alpha e$. Multiplying this expression by 150 yields $150\alpha = 16$.

Substitution is the primary tool for simplifying this integral. Setting $t = \cos x$ leads to $dt = -\sin x \, dx$, which shifts the limits from $0$ to $\pi$ into $1$ and $-1$. Reversing these limits to $[-1, 1]$ absorbs the negative sign, resulting in the expression $I = \int_{-1}^{1} \frac{e^t}{(1+t^2)(e^t+e^{-t})} \, dt$.

Exploiting the symmetry of the interval $[-1, 1]$, the property $\int_{-a}^{a} f(t) \, dt = \int_{-a}^{a} f(-t) \, dt$ shows that $I$ is also equal to $\int_{-1}^{1} \frac{e^{-t}}{(1+t^2)(e^{-t}+e^t)} \, dt$. Combining these two identical forms of $I$ cancels the exponential terms in the numerator and denominator, yielding $2I = \int_{-1}^{1} \frac{1}{1+t^2} \, dt$.

Evaluating this standard integral gives $\tan^{-1} t$ between the bounds of $-1$ and $1$, which produces $\frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$. Dividing by two confirms the final value of $\frac{\pi}{4}$.

When dealing with integrals over a symmetric interval such as $[-2, 2]$, it is often efficient to use the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx$. Applying this to the given expression where $f(x) = \frac{|x^{3+}x|}{e^{x|x|}+1}$, we examine the sum of $f(x)$ and $f(-x)$. Substituting $-x$ into the integrand yields $f(-x) = \frac{|x^{3+}x|}{e^{-x|x|}+1}$ because the absolute value of $-x$ is the same as the absolute value of $x$.

Combining these terms reveals that the sum $f(x) + f(-x)$ simplifies significantly, as $\frac{1}{e^{x|x|}+1} + \frac{1}{e^{-x|x|}+1}$ equals 1. This reduces the original integral to $\int_{0}^{2} |x^{3+}x| \, dx$. Because the expression $x^{3+}x$ is always positive on the interval $[0, 2]$, the absolute value signs are redundant, allowing us to integrate $x^{3+}x$ directly. Calculating the antiderivative gives $\left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_{0}^{2}$, and substituting the limits yields $4 + 2$, which results in a final value of 6.

The most effective strategy for evaluating definite integrals with symmetric limits is the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. Applying this transformation to integrals involving products of polynomials and trigonometric functions often exploits symmetry, causing the non-constant terms in the polynomial to cancel out when summing the original and transformed expressions.

For the integral $I = \int_{0}^{\pi} (\frac{3\pi x^2}{2} - x^3) \frac{\sin x}{1+\cos^2 x} \, dx$, replacing $x$ with $\pi-x$ transforms the polynomial portion $g(x) = \frac{3\pi x^2}{2} - x^3$ into $\frac{\pi^3}{2} - g(x)$, while the trigonometric term $\frac{\sin x}{1+\cos^2 x}$ remains unchanged because $\sin(\pi-x) = \sin x$ and $\cos^2(\pi-x) = \cos^2 x$. Adding the original integral to this transformed version yields $2I = \frac{\pi^3}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} \, dx$. Substituting $u = \cos x$ transforms the remaining integral into $-\int_{1}^{-1} \frac{du}{1+u^2}$, which evaluates to $[\tan^{-1} u]_{-1}^{1} = \frac{\pi}{2}$. Since $2I = \frac{\pi^3}{2} \cdot \frac{\pi}{2} = \frac{\pi^4}{4}$, the value of $I$ is $\frac{\pi^4}{8}$. Multiplying this result by the coefficient $\frac{48}{\pi^4}$ produces 6.

For a function to have a defined limit at a specific point, its left-hand and right-hand limits must be equal. Near $x = -1$, the behavior of the components $\lfloor x \rfloor$ and $\lfloor 2-x \rfloor$ changes. As $x$ approaches $-1$ from the left, $\lfloor x \rfloor = -2$ and $\lfloor 2-x \rfloor = 3$, leading to the limit $a \sin(-\pi) + 3 = 3$. As $x$ approaches $-1$ from the right, $\lfloor x \rfloor = -1$ and $\lfloor 2-x \rfloor = 2$, leading to the limit $a \sin(-\pi/2) + 2 = -a + 2$. Equating these two values, $3 = -a + 2$, establishes that $a = -1$.

With the constant $a$ identified, the function simplifies to $f(x) = -\sin \left( \frac{\pi \lfloor x \rfloor}{2} \right) + \lfloor 2-x \rfloor$. The integral of this function over the interval $[0, 4]$ can be computed by evaluating the integral piece-wise across each unit interval where the floor functions remain constant:

$\int_0^4 \left( -\sin \left( \frac{\pi \lfloor x \rfloor}{2} \right) + \lfloor 2-x \rfloor \right) dx$

For $x \in [0, 1)$, $\lfloor x \rfloor = 0$ and $\lfloor 2-x \rfloor = 1$, so the integral is $\int_0^1 (0 + 1) dx = 1$. For $x \in [1, 2)$, $\lfloor x \rfloor = 1$ and $\lfloor 2-x \rfloor = 0$, resulting in $\int_1^2 (-1 + 0) dx = -1$. For $x \in [2, 3)$, $\lfloor x \rfloor = 2$ and $\lfloor 2-x \rfloor = -1$, yielding $\int_2^3 (0 - 1) dx = -1$. Finally, for $x \in [3, 4)$, $\lfloor x \rfloor = 3$ and $\lfloor 2-x \rfloor = -2$, giving $\int_3^4 (1 - 2) dx = -1$. Summing these values, $1 - 1 - 1 - 1$, yields $-2$.

The integral equation involves a Fredholm structure where the unknown function can be represented as a linear combination of $\sin\theta$ and $\cos\theta$, specifically $f(\theta)=a\sin\theta+b\cos\theta$, because the definite integrals over the fixed interval define constant coefficients. By substituting this form back into the original expression, we can separate the integral into components to identify $a$ and $b$, leading to the relation $f(\theta)=\sin\theta(1+\int_{-\pi/2}^{\pi/2}f(t)dt)+\cos\theta(\int_{-\pi/2}^{\pi/2}tf(t)dt)$. Calculating these integrals over the symmetric interval $[-\pi/2, \pi/2]$ yields the system of equations $a=1+2b$ and $b=2a$, which simplifies to $a=-\frac{1}{3}$ and $b=-\frac{2}{3}$. With the function determined as $f(\theta)=-\frac{1}{3}\sin\theta-\frac{2}{3}\cos\theta$, we compute the definite integral over the specified range:
$\int_{0}^{\pi/2} f(\theta) d\theta = \int_{0}^{\pi/2} \left(-\frac{1}{3}\sin\theta-\frac{2}{3}\cos\theta\right) d\theta$
Evaluating this integral results in $-\frac{1}{3}(1) - \frac{2}{3}(1) = -1$. Taking the absolute value of this outcome yields 1.

To determine the value of the integral for the greatest integer function, identify the intervals where the quadratic expression $-8x^{2+}6x-1$ remains constant by finding the values of $x$ where the function equals integers. Setting the expression equal to $0$ gives $8x^{2-}6x+1=0$, which factors as $(2x-1)(4x-1)=0$, yielding critical points at $x=1/4$ and $x=1/2$. Setting the expression equal to $-1$ results in $8x^{2-}6x=0$, which provides roots at $x=0$ and $x=3/4$. For the expression to equal $-2$, we solve $8x^{2-}6x-1=0$, which provides the root $x=\frac{3+\sqrt{17}}{8}$ within the integration domain, and the expression reaches $-3$ at $x=1$.

Evaluating the greatest integer value within these defined intervals reveals that the expression equals $-1$ on the interval $[0, 1/4]$, $0$ on $[1/4, 1/2]$, $-1$ on $[1/2, 3/4]$, $-2$ on $[3/4, \frac{3+\sqrt{17}}{8}]$, and $-3$ on $[\frac{3+\sqrt{17}}{8}, 1]$. Calculating the definite integral involves summing the areas of these distinct segments:
$\int_{0}^{1/4} -1 dx + \int_{1/4}^{1/2} 0 dx + \int_{1/2}^{3/4} -1 dx + \int_{3/4}^{\frac{3+\sqrt{17}}{8}} -2 dx + \int_{\frac{3+\sqrt{17}}{8}}^{1} -3 dx$
Applying the fundamental theorem of calculus, this summation becomes:
$-[x]_{0}^{1/4} + 0 - [x]_{1/2}^{3/4} - 2[x]_{3/4}^{\frac{3+\sqrt{17}}{8}} - 3[x]_{\frac{3+\sqrt{17}}{8}}^{1}$
Substituting the limits yields the following expression:
$-\left(\frac{1}{4}\right) - \left(\frac{3}{4}-\frac{1}{2}\right) - 2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right) - 3\left(1-\frac{3+\sqrt{17}}{8}\right)$
Simplifying these terms results in:
$-\frac{1}{4} - \frac{1}{4} - 2\left(\frac{\sqrt{17}-3}{8}\right) - 3\left(\frac{5-\sqrt{17}}{8}\right) = -\frac{1}{2} - \frac{\sqrt{17}-3}{4} - \frac{15-3\sqrt{17}}{8}$
Combining these into a single fraction with a denominator of $8$ gives:
$\frac{-4 - 2\sqrt{17} + 6 - 15 + 3\sqrt{17}}{8} = \frac{\sqrt{17}-13}{8}$

Decomposing the integrand into partial fractions allows the expression to be split into two manageable parts:
$\frac{1}{(x^{2-}1)(x^{2-}4)} = \frac{1}{3} \left( \frac{1}{x^{2-}4} - \frac{1}{x^{2-}1} \right)$
Integrating each term using the formula $\int \frac{1}{x^{2-}a^2} dx = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right|$ and multiplying by the factor of 12 yields the indefinite integral $\ln \left| \frac{x-2}{x+2} \right| - 2 \ln \left| \frac{x-1}{x+1} \right|$.

Evaluating this expression across the bounds from 3 to $b$ results in the difference:
$\left( \ln \left( \frac{b-2}{b+2} \right) - 2 \ln \left( \frac{b-1}{b+1} \right) \right) - \left( \ln \left( \frac{1}{5} \right) - 2 \ln \left( \frac{1}{2} \right) \right)$
The lower limit term simplifies to $\ln \left( \frac{1}{5} \cdot 4 \right) = \ln \left( \frac{4}{5} \right)$. Setting the entire integral equal to $\ln \left( \frac{49}{40} \right)$ leads to:
$\ln \left( \frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} \right) = \ln \left( \frac{49}{40} \right) + \ln \left( \frac{4}{5} \right) = \ln \left( \frac{49}{40} \cdot \frac{4}{5} \right) = \ln \left( \frac{49}{50} \right)$
Comparing the arguments inside the logarithms gives $\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} = \frac{49}{50}$. Substituting $b = 6$ into this rational expression satisfies the equality, as $\frac{4}{8} \cdot \frac{7^2}{5^2} = \frac{1}{2} \cdot \frac{49}{25} = \frac{49}{50}$.

The most effective approach for integrating a rational combination of sine and cosine in the denominator is to employ the Weierstrass substitution, defining $t = \tan(x/2)$. Under this transformation, $\sin x$ becomes $\frac{2t}{1+t^2}$ and $\cos x$ becomes $\frac{1-t^2}{1+t^2}$, while the differential element $dx$ is replaced by $\frac{2dt}{1+t^2}$. The integration limits, originally ranging from 0 to $\frac{\pi}{2}$, map to 0 to 1 because $\tan(0)=0$ and $\tan(\pi/4)=1$.

Substituting these into the original integrand simplifies the denominator $3 + 2\sin x + \cos x$ into $3 + \frac{4t}{1+t^2} + \frac{1-t^2}{1+t^2}$, which simplifies algebraically to $\frac{2t^{2+}4t+4}{1+t^2}$. Multiplying this expression by the differential term $\frac{2dt}{1+t^2}$ causes the $(1+t^2)$ factors to cancel out, leaving the integral of $\frac{1}{t^{2+}2t+2}$ with respect to $t$. By completing the square in the denominator, the integrand becomes $\frac{1}{(t+1)^{2+}1}$, which integrates to $\tan^{-1}(t+1)$. Evaluating this antiderivative between the transformed limits of 0 and 1 yields $\tan^{-1}(2) - \tan^{-1}(1)$, which simplifies to $\tan^{-1}(2) - \frac{\pi}{4}$.

The problem hinges on identifying the limit of a Riemann sum as a definite integral, defined as $\lim_{n \to \infty } \frac{1}{n} \sum_{r=1}^n f(\frac{r}{n}) = \int_0^1 f(x) \, dx$. By isolating the term $\frac{(n+1)^{k-1}}{n^{k-1}}$ on the left, we observe it simplifies to $(1+\frac{1}{n})^{k-1}$, which approaches 1 as $n$ goes to infinity. The remaining summation term $\frac{1}{n} \sum_{r=1}^n (k+\frac{r}{n})$ then transforms into the integral $\int_0^1 (k+x) \, dx$, which evaluates to $\left[ kx + \frac{x^2}{2} \right]_0^1 = k + \frac{1}{2}$ or $\frac{2k+1}{2}$.

On the right side, the expression $33 \cdot \lim_{n \to \infty } \frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^k$ converts to $33 \int_0^1 x^k \, dx$, which calculates to $33 \cdot \left[ \frac{x^{k+1}}{k+1} \right]_0^1 = \frac{33}{k+1}$. Equating the two sides results in the expression $\frac{2k+1}{2} = \frac{33}{k+1}$, which rearranges into the quadratic equation:
$2k^2 + 3k - 65 = 0$
Factoring this quadratic gives $(2k+13)(k-5) = 0$, leading to the positive integer solution $k=5$.

The evaluation of the integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{2-x^2}{(2+x^2) \sqrt{4+x^4}} dx$ relies on a strategic substitution to simplify the radical expression and align the integrand with a standard form. Substituting $x = \sqrt{2}t$ implies $dx = \sqrt{2}dt$, which transforms the limits of integration from $0$ to $\sqrt{2}$ into $0$ to $1$. Applying this change yields $\frac{24\sqrt{2}}{\pi} \int_{0}^{1} \frac{2(1-t^2)}{4(1+t^2) \sqrt{1+t^4}} dt$, which simplifies to $\frac{12\sqrt{2}}{\pi} \int_{0}^{1} \frac{1-t^2}{(1+t^2)\sqrt{1+t^4}} dt$.