Indefinite Integrals – JEE Main MathematicsPractice Questions & PYQs

When tackling an integral where an inverse trigonometric function appears in the denominator, identifying a substitution that simplifies the differential is key. Assigning $t = \tan^{-1}(x^{3+}\frac{1}{x^3})$ leads to the differential:
$dt = \frac{1}{1+(x^{3+}\frac{1}{x^3})^2} \cdot \left(3x^2 - \frac{3}{x^4}\right) dx$
Expanding the denominator $1+(x^{3+}\frac{1}{x^3})^2$ results in $1 + x^6 + 2 + \frac{1}{x^6}$, which simplifies cleanly to $\frac{x^{12}+3x^{6+}1}{x^6}$. The term in the numerator, $3x^2 - \frac{3}{x^4}$, can be rewritten as $\frac{3(x^{6-}1)}{x^4}$. Multiplying these components yields $dt = \frac{x^6}{x^{12}+3x^{6+}1} \cdot \frac{3(x^{6-}1)}{x^4} \, dx$, which simplifies to:
$dt = \frac{3(x^{8-}x^2)}{x^{12}+3x^{6+}1} \, dx$
This relationship shows that the entire expression $\frac{(x^{8-}x^2)\,dx}{x^{12}+3x^{6+}1}$ is equivalent to $\frac{1}{3} dt$. Substituting this into the integral produces the simplified form:
$\int \frac{1}{3t} \, dt = \frac{1}{3} \ln|t| + C$
Applying standard logarithm properties, the coefficient $\frac{1}{3}$ becomes an exponent for the argument, resulting in $\ln|t|^{1/3} + C$, which evaluates to:
$\log_e \left( \left| \tan^{-1}\left(x^{3+}\frac{1}{x^3}\right) \right| \right)^{1/3} + C$

Integration by parts is the most effective approach for evaluating integrals of the form $\int \csc^n x \, dx$ by splitting the integrand into $\csc^{n-2} x \cdot \csc^2 x$. This process allows us to construct a reduction formula, linking the original integral to the integral of a lower power of the cosecant function.

To solve $\int \csc^5 x \, dx$, we set $I = \int \csc^3 x \cdot \csc^2 x \, dx$ and designate $u = \csc^3 x$ and $dv = \csc^2 x \, dx$. Applying integration by parts, we determine $du = -3 \csc^3 x \cot x \, dx$ and $v = -\cot x$, leading to $I = -\cot x \csc^3 x - 3 \int \csc^3 x \cot^2 x \, dx$. By replacing $\cot^2 x$ with $\csc^2 x - 1$, the expression becomes $I = -\cot x \csc^3 x - 3 \int \csc^3 x (\csc^2 x - 1) \, dx$, which simplifies to $4I = -\cot x \csc^3 x + 3 \int \csc^3 x \, dx$.

We know the standard integral $\int \csc^3 x \, dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \log_e \left| \tan \frac{x}{2} \right|$. Substituting this result into the previous equation yields $4I = -\cot x \csc^3 x - \frac{3}{2} \csc x \cot x + \frac{3}{2} \log_e \left| \tan \frac{x}{2} \right|$. Factoring out $-\cot x \csc x$ from the first two terms on the right side provides $4I = -\cot x \csc x (\csc^2 x + \frac{3}{2}) + \frac{3}{2} \log_e \left| \tan \frac{x}{2} \right|$. Dividing by 4, we obtain $I = -\frac{1}{4} \cot x \csc x (\csc^2 x + \frac{3}{2}) + \frac{3}{8} \log_e \left| \tan \frac{x}{2} \right| + C$, identifying $\alpha = -1/4$ and $\beta = 3/8$. Therefore, $8(\alpha + \beta) = 8(-\frac{1}{4} + \frac{3}{8}) = 8(\frac{1}{8}) = 1$.

Integrating an expression of the form $\frac{a \cos x + b \sin x}{c \cos x + d \sin x}$ relies on decomposing the numerator into a linear combination of the denominator and its derivative. This technique effectively splits the integral into two standard parts: one involving a constant and another resulting in a logarithmic term.

Converting the integrand $\frac{2-\tan x}{3+\tan x}$ into $\frac{2\cos x - \sin x}{3\cos x + \sin x}$ facilitates this separation. By expressing the numerator as $A(3\cos x + \sin x) + B(\cos x - 3\sin x)$, we create a system of equations by comparing the coefficients of $\sin x$ and $\cos x$. Specifically, matching the coefficients yields $3A + B = 2$ and $A - 3B = -1$. Solving this system leads to $A = \frac{1}{2}$ and $B = \frac{1}{2}$. Substituting these values back into the integral, we evaluate $\int \frac{1}{2} dx + \int \frac{1}{2} \frac{\cos x - 3\sin x}{3\cos x + \sin x} dx$, which results in $\frac{1}{2} x + \frac{1}{2} \ln |3\cos x + \sin x| + C$. Aligning this result with the structure $\frac{1}{2} (\alpha x + \ln |\beta \sin x + \gamma \cos x|) + C$ reveals that $\alpha = 1$, $\beta = 1$, and $\gamma = 3$. Calculating the final value gives $\alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4$.

A substitution of $x^2 = t$ simplifies the integrand dramatically, as the numerator $2x \, dx$ corresponds directly to $dt$. This transforms the integral into $\int \frac{1}{(t+1)(t+3)} \, dt$, which can be resolved through partial fraction decomposition as $\frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} \right) \, dt$. Integrating these terms yields $f(x) = \frac{1}{2} \log_e \left( \frac{x^{2+}1}{x^{2+}3} \right) + C$.

Evaluating this expression at $x=3$ gives $\frac{1}{2} \log_e \left( \frac{10}{12} \right) + C$, which simplifies to $\frac{1}{2} (\log_e 5 - \log_e 6) + C$. Since this matches the given value $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6)$, the integration constant $C$ must be zero. Finding the value at $x=4$ now follows by substituting $x=4$ into the simplified function, resulting in $\frac{1}{2} \log_e \left( \frac{17}{19} \right)$, which is $\frac{1}{2} (\log_e 17 - \log_e 19)$.

To evaluate this integral, we first recognize that the expression can be simplified by identifying a suitable substitution involving the exponential functions. Setting $t = x \ln(\frac{x}{2})$ allows for an elegant reduction of the integrand. Differentiating both sides of this substitution yields $dt = (\ln(\frac{x}{2}) + 1) \, dx$.

The given integral, structured with terms like $(\frac{x}{2})^x$ and $(\frac{2}{x})^x$, effectively takes the form $\int (e^t + e^{-t}) \, dt$ when this substitution is applied. Integrating these exponential terms with respect to $t$ gives $e^t - e^{-t} + C$. Converting back to the original variable by substituting $t = x \ln(\frac{x}{2})$ into the result provides $(\frac{x}{2})^x - (\frac{2}{x})^x + C$, which matches the required form.

To integrate the function $\sqrt{\sec 2x - 1}$, begin by rewriting the integrand as $\sqrt{\frac{1 - \cos 2x}{\cos 2x}}$, which simplifies to $\frac{\sqrt{2} \sin x}{\sqrt{\cos 2x}}$. Using the substitution $\cos x = t$, where $-\sin x \, dx = dt$, the integral becomes $-\sqrt{2} \int \frac{dt}{\sqrt{2t^2 - 1}}$, which equates to $-\ln \left| \cos x + \sqrt{\cos^2 x - 1/2} \right|$. Algebraic rearrangement of this result to fit the logarithmic structure $\alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x(1 + \cos 2x)} \right|$ confirms that $\alpha = -\frac{1}{2}$ and $\beta = \frac{1}{2}$. Calculating $\beta - \alpha$ as $\frac{1}{2} - \left( -\frac{1}{2} \right)$ provides the value 1.

Recognizing the structure of the integrand involves expressing the base terms using natural logarithms, where $(\frac{x}{e})^{2x}$ becomes $\exp(2(x \ln x - x))$ and $(\frac{e}{x})^{2x}$ becomes $\exp(-2(x \ln x - x))$. With the substitution $u = x \ln x - x$, the derivative is $du = \ln x \, dx$, which matches the logarithmic factor in the integral. This transformation leads to the antiderivative of $e^{2u} + e^{-2u}$, resulting in $\frac{1}{2}e^{2u} - \frac{1}{2}e^{-2u} + C$. Substituting $u$ back into the expression yields $\frac{1}{2} (\frac{x}{e})^{2x} - \frac{1}{2} (\frac{e}{x})^{2x} + C$. Aligning this result with the provided form $\frac{1}{\alpha} (\frac{x}{e})^{\beta x} - \frac{1}{\gamma} (\frac{e}{x})^{\delta x} + C$ establishes that $\alpha = 2$, $\beta = 2$, $\gamma = 2$, and $\delta = 2$. Calculating $\alpha + 2\beta + 3\gamma - 4\delta$ produces $2 + 2(2) + 3(2) - 4(2)$, which simplifies to $2 + 4 + 6 - 8$, resulting in $4$.

The integral can be evaluated by recognizing the integrand as the derivative of the product $e^{\sin^2 x} \cos x$. Applying the product rule results in $e^{\sin^2 x}(2 \sin x \cos x)\cos x + e^{\sin^2 x}(-\sin x)$, which simplifies exactly to $e^{\sin^2 x}(\cos x \sin 2x - \sin x)$ since $\sin 2x = 2 \sin x \cos x$.

This identifies the antiderivative as $I(x) = e^{\sin^2 x} \cos x + C$. Applying the condition $I(0) = 1$ gives $1 = e^0(1) + C$, which leads to $1 = 1 + C$ and confirms that $C = 0$. Consequently, evaluating the function at $x = \frac{\pi}{3}$ yields $I\left(\frac{\pi}{3}\right) = e^{\sin^2 (\pi/3)} \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} e^{3/4}$.

Applying integration by parts with $u = x^2$ and $dv = \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$ makes $du = 2x dx$ and $v = -\frac{1}{x \tan x + 1}$. The integration by parts formula $\int u \, dv = uv - \int v \, du$ transforms the expression into $-\frac{x^2}{x \tan x + 1} + \int \frac{2x}{x \tan x + 1} dx$. Expressing the integrand of the remaining integral as $\frac{2x \cos x}{x \sin x + \cos x} dx$ reveals that the numerator is exactly twice the derivative of the denominator, resulting in $2 \ln|x \sin x + \cos x| + C$. Since $I(0) = 0$, the constant $C$ must be $0$. Calculating $I\left(\frac{\pi}{4}\right)$ leads to $-\frac{(\pi/4)^2}{(\pi/4) \tan(\pi/4) + 1} + 2 \ln|(\pi/4) \sin(\pi/4) + \cos(\pi/4)|$, which simplifies to $-\frac{\pi^2}{4(\pi+4)} + \ln\left(\frac{(\pi+4)^2}{32}\right)$.

The evaluation of the integral $f(x)=\int\;\frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$ is most efficiently handled by the substitution $x=\frac{1}{t}$, which implies $dx=-\frac{1}{t^2}dt$. This transformation simplifies the integral into the form $\int\;\frac{-t\,dt}{(3t^{2+}4)\sqrt{4t^{2-}3}}$. Introducing a further substitution $z=\sqrt{4t^{2-}3}$ allows for a cleaner integration, where $z^2=4t^{2-}3$ leads to $t^2=\frac{z^{2+}3}{4}$ and $z\,dz=4t\,dt$, or $t\,dt=\frac{1}{4}z\,dz$. Substituting these relationships reduces the integral to $-\int\;\frac{dz}{3z^{2+}25}$, which can be written as $-\frac{1}{3}\int\;\frac{dz}{z^{2+}(\frac{5}{\sqrt{3}})^2}$.

Applying the standard integral formula $\int \frac{du}{u^{2+}a^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a})$ yields the result $f(x)=-\frac{1}{5\sqrt{3}}\tan^{-1}\left(\frac{\sqrt{3}z}{5}\right)+C$. Converting back to the variable $x$ gives $f(x)=-\frac{1}{5\sqrt{3}}\tan^{-1}\left(\frac{\sqrt{3}}{5}\sqrt{\frac{4-3x^2}{x^2}}\right)+C$. Given the condition $f(0)=0$, we observe the limit of the expression as $x$ approaches zero, where the arctangent term approaches $\frac{\pi}{2}$, determining the constant $C=\frac{\pi}{10\sqrt{3}}$. Evaluating $f(1)$ then results in $-\frac{1}{5\sqrt{3}}\tan^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi}{10\sqrt{3}}$. Using the complementary angle identity $\frac{\pi}{2}-\tan^{-1}(\theta)=\tan^{-1}(\frac{1}{\theta})$, this simplifies to $\frac{1}{5\sqrt{3}}\tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$. Comparing this expression to the form $\frac{1}{\alpha\beta}\tan^{-1}\left(\frac{\alpha}{\beta}\right)$ identifies $\alpha=5$ and $\beta=\sqrt{3}$, leading to the final calculation $\alpha^{2+}\beta^2=25+3=28$.

To evaluate the integral $I(x) = \int \frac{x+1}{x(1+x e^x)^2}\,dx$, we begin by multiplying the numerator and the denominator by $e^x$ to facilitate a substitution. This gives $I(x) = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2}\,dx$. Let $u = x e^x$, then $du = (e^x + x e^x)dx = (1+x)e^x\,dx$. The integral transforms into $\int \frac{du}{u(1+u)^2}$.

Using partial fraction decomposition, we express the integrand as $\frac{1}{u(1+u)^2} = \frac{A}{u} + \frac{B}{1+u} + \frac{C}{(1+u)^2}$. Solving for the coefficients, we find $1 = A(1+u)^2 + Bu(1+u) + Cu$. Setting $u=0$ gives $A=1$; setting $u=-1$ gives $C=-1$; and comparing the coefficients of $u^2$ gives $A+B=0$, so $B=-1$. Thus, the integral becomes $\int \left(\frac{1}{u} - \frac{1}{1+u} - \frac{1}{(1+u)^2}\right)\,du$, which evaluates to $\ln|u| - \ln|1+u| + \frac{1}{1+u} + C$.

Substituting $u = x e^x$ back into the expression, we get $I(x) = \ln|x e^x| - \ln|1+x e^x| + \frac{1}{1+x e^x} + C$, which simplifies to $I(x) = \ln(x) + x - \ln(1+x e^x) + \frac{1}{1+x e^x} + C$. To find the constant $C$, we apply the condition $\lim_{x\to \infty } I(x) = 0$. Using the limit $\lim_{x\to \infty } (\ln(x) + x - \ln(1+x e^x)) = \lim_{x\to \infty } (\ln(x) + x - \ln(x e^x(1+\frac{1}{x e^x}))) = \lim_{x\to \infty } (\ln(x) + x - (\ln(x) + x + \ln(1+\frac{1}{x e^x}))) = 0$, we conclude that $C=0$. Finally, calculating $I(1) = \ln(1) + 1 - \ln(1+e) + \frac{1}{1+e} = 1 + \frac{1}{1+e} - \ln(1+e) = \frac{e+2}{e+1} - \ln(1+e)$.

The integration relies on substituting $x=t^2$ to simplify the term $\sqrt{\frac{x+7}{x}}$, which yields $dx=2t dt$ and transforms the integral into the standard form $2 \int \sqrt{t^{2+}7} dt$. Applying the integral formula $\int \sqrt{t^{2+}a^2} dt = \frac{t}{2}\sqrt{t^{2+}a^2} + \frac{a^2}{2} \log_e|t + \sqrt{t^{2+}a^2}|$ leads to the function $I(x) = \sqrt{x(x+7)} + 7 \log_e(\sqrt{x} + \sqrt{x+7}) + C$.

Substituting $x=9$ into the derived function gives $\sqrt{9(16)} + 7 \log_e(\sqrt{9} + \sqrt{16}) + C = 12 + 7 \log_e 7 + C$. Given that this value equals $12 + 7 \log_e 7$, the constant $C=0$. Evaluating $I(1)$ with this result yields $\sqrt{1(8)} + 7 \log_e(1 + \sqrt{8})$, which simplifies to $\sqrt{8} + 7 \log_e(1 + 2\sqrt{2})$. Comparing this expression with $\alpha + 7 \log_e(1 + 2\sqrt{2})$ reveals that $\alpha = \sqrt{8}$. Squaring this value twice to find $\alpha^4$ results in $8^2$, which is 64.

Evaluating a definite integral defined by a function at a specific point relies on the Fundamental Theorem of Calculus. Because the function is defined by the integral $g(x) = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} \, dx$ and satisfies the condition $g(1)=0$, calculating $g\left(\frac{1}{2}\right)$ is equivalent to evaluating the definite integral $\int_{1}^{1/2} \frac{1}{x} \sqrt{\frac{1-x}{1+x}} \, dx$. This approach transforms the problem into a manageable definite integral over the interval $\left[\frac{1}{2}, 1\right]$, which is best simplified using the trigonometric substitution $x = \cos 2\theta$. Under this substitution, the square root term simplifies to $\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2 \sin^2 \theta}{2 \cos^2 \theta}} = \tan \theta$, and the differential becomes $dx = -2 \sin 2\theta \, d\theta$.

Applying these substitutions shifts the limits of integration from $\left[1, \frac{1}{2}\right]$ to $\left[0, \frac{\pi}{6}\right]$, with the integral expression becoming $\int_{0}^{\pi/6} \frac{1}{\cos 2\theta} \cdot \frac{\sin \theta}{\cos \theta} \cdot (-2 \sin 2\theta) \, d\theta$. By using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, the expression simplifies significantly to $-4 \int_{0}^{\pi/6} \frac{\sin^2 \theta}{\cos 2\theta} \, d\theta$, which can be rewritten as $2 \int_{0}^{\pi/6} (1-\sec 2\theta) \, d\theta$ by utilizing the identity $2\sin^2 \theta = 1-\cos 2\theta$. Evaluating this integral yields $\left[ 2\theta - \ln|\sec 2\theta + \tan 2\theta| \right]_{0}^{\pi/6}$, resulting in $\frac{\pi}{3} - \ln|2 + \sqrt{3}|$. Since $\ln(2 + \sqrt{3})^{-1} = \ln(2 - \sqrt{3})$, which is equivalent to $\log_e \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right)$, the final value is $\log_e \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right) + \frac{\pi}{3}$.

The function $I(x)$ originates from the integration of a specific derivative structure, which can be identified by differentiating $\frac{\tan x}{\sin^{2022} x}$. Applying the quotient rule, the derivative is $\frac{\sec^2 x \sin^{2022} x - 2022 \sin^{2021} x \cos x \tan x}{\sin^{4044} x}$, which reduces to $\frac{\sec^2 x - 2022}{\sin^{2022} x}$. Therefore, the integral evaluates to $I(x) = \frac{\tan x}{\sin^{2022} x} + C$.

Using the condition $I(\pi/4) = 2^{1011}$, we can solve for the integration constant $C$. Substituting the values, we find $\frac{\tan(\pi/4)}{(\sin(\pi/4))^{2022}} + C = 2^{1011}$. Since $(\sin(\pi/4))^{2022}$ equals $(1/\sqrt{2})^{2022}$, which is $1/2^{1011}$, the expression simplifies to $2^{1011} + C = 2^{1011}$, proving that $C = 0$. The function is thus defined as $I(x) = \frac{\tan x}{\sin^{2022} x}$.

Calculating $I(\pi/3)$ and $I(\pi/6)$ allows us to verify the required relationship. For $x = \pi/3$, $I(\pi/3) = \frac{\tan(\pi/3)}{(\sin(\pi/3))^{2022}} = \frac{\sqrt{3}}{(\sqrt{3}/2)^{2022}} = \frac{\sqrt{3} \cdot 2^{2022}}{3^{1011}}$. For $x = \pi/6$, $I(\pi/6) = \frac{\tan(\pi/6)}{(\sin(\pi/6))^{2022}} = \frac{1/\sqrt{3}}{(1/2)^{2022}} = \frac{2^{2022}}{\sqrt{3}}$. Substituting these into the first option, we compute $3^{1010} I(\pi/3) - I(\pi/6) = 3^{1010} \cdot \frac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} - \frac{2^{2022}}{\sqrt{3}}$. This expression simplifies to $\frac{\sqrt{3} \cdot 2^{2022}}{3} - \frac{2^{2022}}{\sqrt{3}}$, which results in $\frac{2^{2022}}{\sqrt{3}} - \frac{2^{2022}}{\sqrt{3}} = 0$.

To solve the integral $\int \frac{(1-1/\sqrt{3})(\cos x - \sin x)}{1 + (2/\sqrt{3})\sin 2x} \, dx$, we first simplify the integrand by factoring out constants and applying trigonometric identities. Multiplying the numerator and denominator by $\sqrt{3}$, we can rewrite the denominator $1 + \frac{2}{\sqrt{3}}\sin 2x$ as $\frac{2}{\sqrt{3}} \left( \frac{\sqrt{3}}{2} + \sin 2x \right)$. Recognizing that $\frac{\sqrt{3}}{2} = \sin \frac{\pi}{3}$, the denominator becomes $\frac{2}{\sqrt{3}} \left( \sin \frac{\pi}{3} + \sin 2x \right)$.