Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Rationalizing the integrand is the most direct strategy when dealing with a sum of square roots in the denominator. By multiplying the numerator and denominator by the conjugate $\sqrt{3+x}-\sqrt{1+x}$, the denominator simplifies to the constant $(3+x)-(1+x)=2$. This transformation allows the integral to be expressed as:

$\frac{1}{2} \int_{0}^{1} (\sqrt{3+x} - \sqrt{1+x}) \, dx$

Evaluating this expression involves applying the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$ to each term, resulting in $\frac{1}{3} [(3+x)^{3/2} - (1+x)^{3/2}]$ over the interval from 0 to 1. Substituting the upper limit $x=1$ yields $8 - 2\sqrt{2}$, while the lower limit $x=0$ results in $3\sqrt{3} - 1$. Taking the difference between these values produces $3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$. Comparing this result to the form $a+b\sqrt{2}+c\sqrt{3}$ identifies the coefficients as $a=3$, $b=-\frac{2}{3}$, and $c=-1$. Computing the final required expression $2a+3b-4c$ with these values yields $2(3) + 3(-\frac{2}{3}) - 4(-1) = 6 - 2 + 4$, which equals 8.

To determine the unknown constants, we translate the provided conditions into a system of linear equations involving $a$, $b$, and $c$. Starting with the value of the function at zero, we find $f(0) = a(e^0)^2 + b(e^0) + c(0) = a + b$, which gives the equation $a + b = -1$.

Next, we compute the derivative $f'(x) = 2a e^{2x} + b e^x + c$. Substituting the given derivative at $x = \ln 2$, we utilize the properties $e^{\ln 2} = 2$ and $e^{2\ln 2} = 4$ to obtain $2a(4) + b(2) + c = 21$, which simplifies to $8a + 2b + c = 21$.

The third condition involves the definite integral $\int_0^{\ln 4} (a e^{2x} + b e^x) dx = \frac{39}{2}$. Integrating this yields $\left[ \frac{a}{2} e^{2x} + b e^x \right]_0^{\ln 4} = \left( \frac{a}{2} (16) + b(4) \right) - \left( \frac{a}{2} + b \right) = 7.5a + 3b$. Equating this expression to $\frac{39}{2}$ results in $15a + 6b = 39$.

With two equations involving $a$ and $b$, specifically $a + b = -1$ and $15a + 6b = 39$, we solve for the variables by substituting $b = -a - 1$ into the second equation to get $15a + 6(-a - 1) = 39$. This leads to $9a = 45$, so $a = 5$, and consequently $b = -6$. Using the derivative equation $8(5) + 2(-6) + c = 21$, we find $40 - 12 + c = 21$, which reveals $c = -7$. The final value is calculated as $|a + b + c| = |5 - 6 - 7| = |-8| = 8$.

The limit of the given summation as $n \to \infty$ converges to a definite integral by treating the sum as a Riemann sum where the step size is $1/n$ and the ratio $r/n$ approaches the continuous variable $x$ over the interval $[0, 1]$. By expressing each term of the series in terms of $r/n$, the limit transforms into the definite integral $\int_0^1 \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx$. Rearranging this integrand by dividing both the numerator and the denominator by $x^2$ allows it to be expressed in the form $\int_0^1 \frac{1/x^2 - 1}{(x+1/x)\sqrt{x^{2+}1/x^2}} dx$.

Substituting $t = x + 1/x$, which implies $dt = (1 - 1/x^2) dx$, changes the integral limits from $[0, 1]$ to $[\infty , 2]$ and results in the expression $-\int_\infty ^2 \frac{dt}{t\sqrt{t^{2-}2}}$. Introducing a further substitution $t^2 - 2 = \alpha^2$, where $t dt = \alpha d\alpha$, transforms this into $-\int_\infty ^{\sqrt{2}} \frac{d\alpha}{\alpha^2 + 2}$, which integrates to $-\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\alpha}{\sqrt{2}}\right)$ evaluated from $\infty$ to $\sqrt{2}$. This calculation yields $\frac{1}{\sqrt{2}} \left( \frac{\pi}{2} - \frac{\pi}{4} \right)$, which simplifies to $\frac{\pi}{4\sqrt{2}}$. Comparing this result to $\frac{\pi}{k}$, we identify $k = 4\sqrt{2}$, which means $k^2 = 32$.

The evaluation begins by simplifying the integral through substitution, setting $\cos x = t^2$, which implies $\sin x \, dx = -2t \, dt$. As the limits of integration change from $0$ and $\frac{\pi}{2}$ to $1$ and $0$, and noting $\sin 2x = 2 \sin x \cos x$, the expression transforms into $4 \int_0^1 t^{14} \sqrt{1+t^5} \, dt$.

To resolve the radical, a second substitution is applied using $1 + t^5 = k^2$, which results in $5t^4 \, dt = 2k \, dk$. This converts the integrand into $\frac{8}{5} \int_1^{\sqrt{2}} (k^6 - 2k^4 + k^2) \, dk$, allowing for direct integration.

Evaluating the definite integral leads to $\frac{8}{5} \left[ \frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3} \right]_1^{\sqrt{2}}$, which simplifies to $\frac{176\sqrt{2} - 64}{525}$. Since the original integral is equated to $\frac{n \sqrt{2} - 64}{525}$, we identify $n = 176$.

The slope of the line $45x+5y+3=0$ is determined by rewriting the equation as $y=-9x-\frac{3}{5}$, which reveals a slope of $-9$. Setting this value equal to $27r_1+\frac{9r_2}{2}$ establishes the relationship $27r_1+\frac{9r_2}{2}=-9$.

Since the limit $\lim_{x \to 3}\left(\frac{\int_{3}^{x} 8 t^2 \, dt}{\frac{3 r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3 x}\right)$ takes the indeterminate form of $\frac{0}{0}$ at $x=3$, L'Hôpital's rule is applied by differentiating both the numerator and the denominator with respect to $x$. Using the Leibniz integral rule, the derivative of the numerator $\int_{3}^{x} 8t^2 \, dt$ is $8x^2$, and the derivative of the denominator $\frac{3r_2x}{2} - r_2x^2 - r_1x^3 - 3x$ is $\frac{3r_2}{2}-2r_2x-3r_1x^{2-}3$. Evaluating these derivatives at $x=3$ yields a numerator of $8(3^2)=72$ and a denominator of $\frac{3r_2}{2}-6r_2-27r_1-3$. Factoring the denominator results in $-\left(27r_1+\frac{9r_2}{2}\right)-3$. Substituting the value of $-9$ from the slope relationship gives $9-3=6$, and dividing $72$ by $6$ results in $12$.

Computing the integral of $\cos^4 x$ over the interval $[0, \frac{\pi}{3}]$ relies on using power-reduction trigonometric identities to express the integrand as a sum of simple cosine terms. By substituting $\cos^2 x = \frac{1+\cos 2x}{2}$, the integrand becomes $(\frac{1+\cos 2x}{2})^2$, which expands to $\frac{1}{4}(1 + 2\cos 2x + \cos^2 2x)$. Applying the identity again to $\cos^2 2x$ as $\frac{1+\cos 4x}{2}$ allows the expression to be rewritten as $\frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{4}(\frac{1+\cos 4x}{2})$, which simplifies to $\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$. Integrating this term by term over the interval gives the following:

$\int_{0}^{\frac{\pi}{3}} \left( \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \right) dx = \left[ \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x \right]_{0}^{\frac{\pi}{3}}$

Evaluating this expression at the boundaries, the first term results in $\frac{3}{8}(\frac{\pi}{3}) = \frac{\pi}{8}$. The sine terms contribute $\frac{1}{4}\sin\frac{2\pi}{3} + \frac{1}{32}\sin\frac{4\pi}{3}$, which evaluates to $\frac{1}{4}(\frac{\sqrt{3}}{2}) + \frac{1}{32}(-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} = \frac{7\sqrt{3}}{64}$. Comparing the resulting $\frac{\pi}{8} + \frac{7\sqrt{3}}{64}$ to $a \pi + b \sqrt{3}$ identifies $a = \frac{1}{8}$ and $b = \frac{7}{64}$. Consequently, calculating $9a+8b$ yields $9(\frac{1}{8}) + 8(\frac{7}{64}) = \frac{9}{8} + \frac{7}{8} = 2$.

The integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{(1+e^{\sin x})(1+\sin^4 x)} \, dx$ is solved effectively using the King's property, which states that $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. Applying this by replacing $x$ with $-x$ results in an equivalent integral where the denominator's exponential term transforms into $\frac{e^{\sin x}}{1+e^{\sin x}}$. Adding this transformed integral to the original causes the exponential components to combine and cancel out, simplifying the expression to $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \, dx$. Since the integrand is an even function, we can simplify this further to $I = \int_{0}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \, dx$.

Substituting $\sin x = t$ leads to $dt = \cos x \, dx$, converting the integral to $I = \int_{0}^{1} \frac{8\sqrt{2}}{1+t^4} \, dt$. To evaluate this, the integrand is split into two parts: $\frac{4\sqrt{2}(1+1/t^2)}{t^{2+}1/t^2}$ and $\frac{4\sqrt{2}(1-1/t^2)}{t^{2+}1/t^2}$. These result in the forms $\int \frac{dz}{z^{2+}2}$ and $\int \frac{dk}{k^{2-}2}$ after making the substitutions $z = t - 1/t$ and $k = t + 1/t$, respectively. Integrating these expressions over their corresponding bounds yields $2\pi$ from the inverse tangent term and $2 \log_e(3+2\sqrt{2})$ from the logarithmic term after rationalizing the arguments. Matching these values to the expression $\alpha \pi + \beta \log_e(3+2\sqrt{2})$, we find $\alpha = 2$ and $\beta = 2$. Consequently, calculating $\alpha^2 + \beta^2$ gives $4 + 4 = 8$.

The shortest distance between two skew lines is determined by the projection of the vector connecting a point on each line onto the common perpendicular vector, which is found via the cross product of the two lines' direction vectors. For the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$, the direction vectors are $\langle 2, 3, 4 \rangle$ and $\langle 1, -3, 2 \rangle$. Their cross product results in the vector $\langle 18, 0, -9 \rangle$, which has a magnitude of $\sqrt{18^2 + (-9)^2} = \sqrt{324 + 81} = 9\sqrt{5}$. Taking the point $(-2, -3, 5)$ from the first line and $(3, 2, -4)$ from the second, the vector joining them is $\langle 5, 5, -9 \rangle$. The absolute value of the dot product between this vector and the cross product vector is $|5(18) + 5(0) - 9(-9)| = |90 + 81| = 171$. Dividing this by the magnitude of the cross product gives a shortest distance of $\frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}}$.

Equating this calculated distance to the given expression $\frac{38}{3\sqrt{5}} k$ yields $\frac{19}{\sqrt{5}} = \frac{38}{3\sqrt{5}} k$. Cancelling the common terms, we find $19 = \frac{38}{3} k$, which simplifies to $k = \frac{19 \times 3}{38} = \frac{3}{2}$. We now evaluate the definite integral \int_{0}^{3/2} \[x^2]$ dx$by partitioning the interval based on the jump discontinuities of the greatest integer function$[x^2]$. The value of$x^2$is 1 at$x=1$and 2 at$x=\sqrt{2}$, so the integral becomes$\int_{0}^{1} 0 , dx + \int_{1}^{\sqrt{2}} 1 , dx + \int_{\sqrt{2}}^{3/2} 2 , dx$.

Calculating these sections individually results in $0 + (\sqrt{2} - 1) + 2(\frac{3}{2} - \sqrt{2})$, which simplifies to $\sqrt{2} - 1 + 3 - 2\sqrt{2} = 2 - \sqrt{2}$. Comparing this to the given form $\alpha - \sqrt{\alpha}$, it is evident that $\alpha = 2$. Finally, substituting $\alpha$ into the expression $6\alpha^3$ leads to $6(2)^3 = 6 \times 8 = 48$.

To evaluate integrals involving powers of binomials like $(1-x^k)^n$ between zero and one, the most effective approach is integration by parts. By choosing the binomial term as the function to differentiate and the differential element $dx$ as the function to integrate, the resulting boundary terms vanish when evaluated at the limits of zero and one. This process reveals a direct recurrence relation where $I_n = \frac{nk}{nk+1} I_{n-1}$, establishing a consistent ratio between consecutive integrals.

Applying this relation for $n=21$ yields the ratio $\frac{I_{21}}{I_{20}} = \frac{21k}{21k+1}$. Given the condition $147 I_{20} = 148 I_{21}$, we can rewrite the equation as $\frac{I_{21}}{I_{20}} = \frac{147}{148}$. Equating the two expressions for the ratio, $\frac{21k}{21k+1} = \frac{147}{148}$, leads to the equation $21k \cdot 148 = 147(21k + 1)$. Dividing both sides by 21 simplifies the expression to $148k = 7(21k + 1)$, which further reduces to $148k = 147k + 7$, confirming that $k=7$.

The derivative of the function $f(x)$ is determined by the Fundamental Theorem of Calculus, which gives $f'(x) = (e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}$. Local extrema occur at the critical points where $f'(x) = 0$, specifically $x = 0, 1/2, 2, 3, 5$. Evaluating the sign changes of $f'(x)$ across these intervals reveals that the function transitions between positive and negative values at $x = 0, 1/2, 2,$ and $5$ due to their odd exponents, whereas the even exponent of $(x-3)^{12}$ prevents a sign change at $x = 3$.

By observing these transitions, $f(x)$ exhibits local maxima where the derivative changes from positive to negative, occurring at $x = 0$ and $x = 2$, and local minima where the derivative changes from negative to positive, occurring at $x = 1/2$ and $x = 5$. Calculating the sum of the squares for the maxima yields $p = 0^2 + 2^2 = 4$, and the sum of the values for the minima yields $q = 1/2 + 5 = 11/2$. Substituting these into the required expression results in $p^2 + 2q = 4^2 + 2(11/2) = 16 + 11 = 27$.

The Beta function defined as $\beta(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$ provides a standard framework for evaluating integrals that involve powers of a variable and its linear complement. By applying the substitution $t = x^{10}$, we find that $x = t^{1/10}$ and $dx = \frac{1}{10} t^{-9/10} dt$.

This transformation changes the original integral into $\frac{1}{10} \int_0^1 t^{-9/10}(1-t)^{20} dt$. Comparing this structure to the definition of the Beta function allows for the identification of parameters, where $m-1 = -9/10$ leads to $b = 1/10$, $n-1 = 20$ leads to $c = 21$, and the coefficient $a = 1/10$. Summing these values gives $a+b+c = \frac{1}{10} + \frac{1}{10} + 21 = 21.2$, and multiplying by 100 yields a final result of 2120.

To evaluate this definite integral, we use the double-angle identity for cosine, $\cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$. Recognizing that the inverse trigonometric expression $\cot^{-1} \sqrt{\frac{1-x}{1+x}}$ is equivalent to $\tan^{-1} \sqrt{\frac{1+x}{1-x}}$, we can define $\theta = \tan^{-1} \sqrt{\frac{1+x}{1-x}}$, which makes $\tan \theta = \sqrt{\frac{1+x}{1-x}}$. Applying this identity transforms the integrand into a much simpler algebraic expression.

Substituting our substitution into the cosine identity gives:
$\frac{1 - \left(\sqrt{\frac{1+x}{1-x}}\right)^2}{1 + \left(\sqrt{\frac{1+x}{1-x}}\right)^2} = \frac{1 - \frac{1+x}{1-x}}{1 + \frac{1+x}{1-x}}$
Multiplying both the numerator and the denominator by $(1-x)$ simplifies the expression to $\frac{(1-x) - (1+x)}{(1-x) + (1+x)}$, which reduces to $\frac{-2x}{2}$, or simply $-x$. The integral is then computed over the range from $1/4$ to $3/4$:
$\int_{1/4}^{3/4} (-x) \, dx = - \left[ \frac{x^2}{2} \right]_{1/4}^{3/4}$
Evaluating this expression at the limits yields $-\frac{1}{2} \left( \frac{9}{16} - \frac{1}{16} \right)$, which simplifies to $-\frac{1}{2} \left( \frac{8}{16} \right)$, resulting in an final value of $-\frac{1}{4}$.

For a piecewise function to be differentiable at the transition point, it must first be continuous, meaning the two branches must evaluate to the same value at $x=1$. Equating the two expressions at this point yields $1^{2+}3(1)+a = b(1)+2$, which simplifies to $4+a = b+2$, or $a = b-2$. Because the function is differentiable, the derivative must also be continuous at $x=1$. The derivative of the first piece is $2x+3$, and the derivative of the second piece is $b$, so setting these equal at $x=1$ gives $2(1)+3 = b$, which means $b=5$. Substituting this value back into the continuity equation results in $a = 5-2 = 3$.

With the constants determined, the function becomes $f(x)=x^{2+}3x+3$ for $x \le 1$ and $f(x)=5x+2$ for $x > 1$. To compute $\int_{-2}^{2} f(x) dx$, we split the integral at $x=1$. The first part is $\int_{-2}^1 (x^{2+}3x+3) dx$, which evaluates to $\left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{-2}^1 = \left( \frac{1}{3} + \frac{3}{2} + 3 \right) - \left( -\frac{8}{3} + 6 - 6 \right) = 7.5$. The second part is $\int_1^2 (5x+2) dx$, which evaluates to $\left[ \frac{5x^2}{2} + 2x \right]_1^2 = (10 + 4) - (\frac{5}{2} + 2) = 14 - 4.5 = 9.5$. Adding these two segments together, $7.5 + 9.5$ results in $17$.

Splitting the initial integral $\int_{0}^{\pi/2} f(\sin 2x) \sin x \, dx$ into two separate intervals, from $0$ to $\frac{\pi}{4}$ and from $\frac{\pi}{4}$ to $\frac{\pi}{2}$, allows for a simplification based on symmetry. Applying the substitution $x = \frac{\pi}{2} - t$ to the second part transforms it into $\int_{0}^{\pi/4} f(\cos 2t) \cos t \, dt$, effectively aligning the trigonometric arguments across the domain. By combining these terms with the second integral $\alpha \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx$, we derive the expression $\int_{0}^{\pi/4} f(\cos 2x) (\sqrt{2} + \alpha) \cos x \, dx = 0$. Since the integral of the function itself is non-vanishing, the term $(\sqrt{2} + \alpha)$ must necessarily equal zero, identifying $\alpha$ as $-\sqrt{2}$.

The interval $1 \leq x \leq 2$ simplifies the expression significantly because the greatest integer function $[x]$ consistently evaluates to $1$ throughout this range. This allows the integrand e^{[x] + \[x^3]$}$to be written as$e^{1 + $[x^3]$}$, which is equal to$e \cdot e^{[x^3]$}$. Consequently, the integral becomes$e \int_{1}^{2} x^2 e^{[x^3]$} dx$.

To resolve this integral, applying the substitution $t = x^3$ is the most direct approach. Differentiating both sides gives $dt = 3x^2 dx$, which means $x^2 dx = \frac{1}{3} dt$. With this change of variables, the limits of integration for $t$ become $1^3 = 1$ and $2^3 = 8$. Substituting these into the expression results in $\frac{e}{3} \int_{1}^{8} e^{[t]} dt$.

The integral $\int_{1}^{8} e^{[t]} dt$ is the sum of integrals across consecutive integer segments: $\int_{1}^{2} e^1 dt + \int_{2}^{3} e^2 dt + \dots + \int_{7}^{8} e^7 dt$. Because each interval has a width of $1$, this simplifies to the sum $e^1 + e^2 + \dots + e^7$. This is a geometric progression with a sum of $e \frac{e^7 - 1}{e - 1}$.

Multiplying this result by the coefficient $\frac{3(e - 1)}{e}$ produces a clean cancellation of terms. Specifically, $\frac{3(e - 1)}{e} \cdot \frac{e}{3} \cdot \frac{e(e^7 - 1)}{e - 1}$ reduces to $e(e^7 - 1)$, which simplifies to $e^8 - e$.

The first statement requires evaluating the limit of the sum of the first $n$ even integers divided by $n^2$. Since the sum $2+4+6+\cdots+2n$ is equivalent to $n(n+1)$, the fraction simplifies to $\frac{n^{2+}n}{n^2}$, which approaches 1 as $n$ tends to infinity. The second statement is evaluated using the definite integral definition of a Riemann sum, where the term $\frac{1}{n^{16}} \sum_{r=1}^n r^{15}$ is rewritten as $\frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^{15}$. As $n$ approaches infinity, this summation converges to the integral $\int_{0}^{1} x^{15} \, dx$, which calculates to $\frac{1}{16}$, confirming that both assertions are mathematically valid.

The behavior of this product as $n \to \infty$ is best understood through the Sandwich Theorem, which allows us to bound the expression between its smallest and largest terms raised to the power of $n$. Each factor $(2^{1/2}-2^{1/k})$ is a positive value strictly between 0 and 1. By establishing that $(2^{1/2}-2^{1/3})^n \le P \le (2^{1/2}-2^{1/(2n+1)})^n$, we observe that as $n$ grows, both the lower and upper bounds approach 0 because they represent a fraction less than 1 being raised to an increasingly large power. Consequently, the limit of the entire product must converge to 0.

Definite integrals involving functions defined over ranges where the variable can be shifted by the sum of the limits often simplify through the property that $\int_{0}^{a} g(x) \, dx = \int_{0}^{a} g(a - x) \, dx$. Defining the integral as $I = \int_{0}^{\pi} f(x) \sin x \, dx$, this property allows the integral to be rewritten as $I = \int_{0}^{\pi} f(\pi - x) \sin(\pi - x) \, dx$. Because $\sin(\pi - x)$ is identical to $\sin x$, this equality simplifies to $I = \int_{0}^{\pi} f(\pi - x) \sin x \, dx$.

Summing the two equivalent expressions for $I$ yields $2I = \int_{0}^{\pi} (f(x) + f(\pi - x)) \sin x \, dx$. Substituting the given relation $f(x) + f(\pi - x) = \pi^2$ into this expression results in $2I = \pi^2 \int_{0}^{\pi} \sin x \, dx$. Evaluating the integral of $\sin x$ over the interval $[0, \pi]$ gives $2$, which leads to $2I = 2\pi^2$. Dividing by two confirms that $I = \pi^2$.

Recognizing that the denominator factors into $(t^{2+}1)(t^{4-}t^{2+}1)$ provides the key to simplifying the expression $\frac{t^{4+}1}{t^{6+}1}$. By partitioning the numerator as $(t^{4-}t^{2+}1) + t^2$, the integrand separates into the sum of two distinct fractions: $\frac{1}{t^{2+}1}$ and $\frac{t^2}{t^{6+}1}$. Integration of the first term results in the standard inverse tangent function $\tan^{-1}(t)$. For the second term, observe that the derivative of $t^3$ is $3t^2$, which allows the expression to be rewritten as $\frac{1}{3} \frac{3t^2}{(t^3)^{2+}1} dt$. This leads to the antiderivative $\frac{1}{3} \tan^{-1}(t^3)$.