Indefinite Integrals – JEE Main MathematicsPractice Questions & PYQs

The integral $\int \frac{3\,dx}{(4x+6)\sqrt{4x^{2+}8x+3}}$ is evaluated by factoring the quadratic expression $4x^{2+}8x+3$ into $(2x+1)(2x+3)$ and extracting a factor of 2 from the linear term, yielding $\int \frac{3\,dx}{2(2x+3)\sqrt{(2x+1)(2x+3)}}$. Applying the substitution $2x+3 = \frac{1}{t}$ transforms the differential $dx$ into $-\frac{1}{2t^2} dt$, and the integrand becomes $\int \frac{-3/2}{t^2 \cdot (1/t) \sqrt{(1-2t)/t \cdot 1/t}} dt$, which simplifies to $-\frac{3}{4} \int (1-2t)^{-1/2} dt$. Integrating this gives $I(x) = \frac{3}{4}\sqrt{1-2t} + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C$. Given $I(0) = \frac{3}{4\sqrt{3}} + C = \frac{\sqrt{3}}{4} + 20$, the constant $C$ is determined to be $20$. Substituting $x=\frac{1}{2}$ into the result provides $I\left(\frac{1}{2}\right) = \frac{3}{4} \sqrt{\frac{1+1}{1+3}} + 20 = \frac{3}{4} \sqrt{\frac{2}{4}} + 20 = \frac{3\sqrt{2}}{8} + 20$. With $a=3$, $b=8$, and $c=20$, the sum $a+b+c = 3+8+20 = 31$.

Substituting $x = \ln t$ transforms the differential $dt = e^x dx$ and the integral into $f(t) = \int e^x \frac{1 - \sin x}{1 - \cos x} dx$. Applying the half-angle identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1 - \cos x = 2 \sin^2(x/2)$, the integrand simplifies to $e^x (\frac{1}{2} \csc^2(x/2) - \cot(x/2))$. Recognizing this expression as the form $\int e^x (g'(x) + g(x)) dx = e^x g(x) + C$ where $g(x) = -\cot(x/2)$, the antiderivative becomes $f(t) = -t \cot(\frac{\ln t}{2}) + C$. Substituting the condition $f(e^{\pi/2}) = -e^{\pi/2}$ yields $-e^{\pi/2} \cot(\pi/4) + C = -e^{\pi/2}$, which simplifies to $-e^{\pi/2}(1) + C = -e^{\pi/2}$ and confirms $C = 0$. Consequently, $f(e^{\pi/4}) = -e^{\pi/4} \cot(\pi/8)$, and since $\cot(\pi/8) = \sqrt{2} + 1$, the expression becomes $f(e^{\pi/4}) = -e^{\pi/4}(\sqrt{2} + 1)$, identifying $\alpha = -1 - \sqrt{2}$.

Substituting $x = t^6$ transforms the integral, where $dx = 6t^5 dt$, $x^{2/3} = t^4$, and $x^{1/2} = t^3$. The integral becomes:
$\int \frac{6t^5 dt}{t^4 + 2t^3} = \int \frac{6t^2}{t + 2} dt = 6 \int \left(t - 2 + \frac{4}{t+2}\right) dt$
Integrating this expression yields $f(x) = 3t^2 - 12t + 24\ln|t+2| + C = 3x^{1/3} - 12x^{1/6} + 24\ln|x^{1/6} + 2| + C$. Applying the initial condition $f(0) = -26 + 24\ln(2)$ gives $3(0) - 12(0) + 24\ln(2) + C = -26 + 24\ln(2)$, confirming $C = -26$. Substituting $x = 1$ into the derived function results in $f(1) = 3(1) - 12(1) + 24\ln(1 + 2) - 26 = 3 - 12 - 26 + 24\ln(3) = -35 + 24\ln(3)$. Identifying $a = -35$ and $b = 24$, the sum $a + b$ is $-35 + 24 = -11$.

The integral is of the form $\int e^x (g(x) + g'(x)) dx$, which evaluates to $e^x g(x) + C$. Rewriting the integrand $\frac{(2-x^2)e^x}{\sqrt{1+x}(1-x)^{3/2}}$ as $e^x \left( \sqrt{\frac{1+x}{1-x}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}} \right)$ identifies the function $g(x) = \sqrt{\frac{1+x}{1-x}}$ and its derivative $g'(x) = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$.

Consequently, the function is $f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C$. Given the condition $f(0) = 0$, substituting $x=0$ into the equation yields $e^0 \sqrt{\frac{1+0}{1-0}} + C = 1(1) + C = 0$, which determines the constant $C = -1$. Evaluating the resulting function $f(x) = e^x \sqrt{\frac{1+x}{1-x}} - 1$ at $x = \frac{1}{2}$ gives $f\left(\frac{1}{2}\right) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} - 1 = \sqrt{e} \cdot \sqrt{3} - 1 = \sqrt{3e} - 1$.

Rewriting the integrand as $(\cot x)^{-5/2} \csc^6 x \, dx$ and substituting $u = \cot x$, where $du = -\csc^2 x \, dx$, transforms the integral into $-\int u^{-5/2} (1+u^2)^3 \, du$. Expanding the polynomial $(1+u^2)^3 = 1 + 3u^2 + 3u^4 + u^6$ and integrating term by term results in $-\int (u^{-5/2} + 3u^{-1/2} + 3u^{3/2} + u^{7/2}) \, du$, which evaluates to $\frac{2}{3} u^{-3/2} - 6 u^{1/2} - \frac{6}{5} u^{5/2} - \frac{2}{9} u^{9/2} + C$. Comparing these coefficients to the provided form yields $p_1=2, q_1=9, p_2=6, q_2=5, p_3=6, q_3=1, p_4=2, q_4=3$. Calculating the final value $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ gives $15 \cdot \frac{2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 15 \cdot \frac{144}{135} = 16$.

The integrand $\frac{1-5\cos^2 x}{\sin^5 x \cos^2 x}$ corresponds to the derivative of the function $f(x) = \frac{1}{\sin^4 x \cos x} = \csc^4 x \sec x$. Evaluating this function at the given upper and lower limits, we find $f\left(\frac{\pi}{6}\right) = \frac{1}{\sin^4\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{6}\right)} = \frac{1}{\left(\frac{1}{2}\right)^4 \left(\frac{\sqrt{3}}{2}\right)} = \frac{1}{\frac{\sqrt{3}}{32}} = \frac{32}{\sqrt{3}}$.

Similarly, $f\left(\frac{\pi}{4}\right) = \frac{1}{\sin^4\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right)} = \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4 \left(\frac{1}{\sqrt{2}}\right)} = \frac{1}{\left(\frac{1}{4}\right) \left(\frac{1}{\sqrt{2}}\right)} = 4\sqrt{2}$. The difference $f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right)$ is then $\frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{32 - 4\sqrt{6}}{\sqrt{3}} = \frac{4}{\sqrt{3}}(8-\sqrt{6})$.

Integration of rational functions typically relies on splitting the integrand into manageable parts, often using the derivative of the denominator to facilitate logarithmic substitution and partial fractions to address remaining terms. Recognizing that the derivative of $x^{2}+2x-15$ is $2x+2$, the numerator $16x+24$ can be expressed as $8(2x+2)+8$, transforming the integral into $8 \int \frac{2x+2}{x^{2}+2x-15} dx + 8 \int \frac{1}{(x+5)(x-3)} dx$. The first part yields $8 \ln |x^{2}+2x-15|$, and utilizing the partial fraction decomposition $\frac{1}{(x+5)(x-3)} = \frac{1}{8}(\frac{1}{x-3} - \frac{1}{x+5})$ allows the second part to integrate into $\ln \left| \frac{x-3}{x+5} \right|$, leading to the general form $f(x) = 8 \ln |x^{2}+2x-15| + \ln \left| \frac{x-3}{x+5} \right| + C$. Substituting $x=4$ results in $8 \ln |16+8-15| + \ln |1/9| + C$, which simplifies to $16 \ln 3 - 2 \ln 3 + C$, or $14 \ln 3 + C$; since $f(4)=14 \ln 3$, the constant $C$ must be $0$. Finally, evaluating at $x=7$ gives $f(7) = 8 \ln |49+14-15| + \ln |4/12| = 8 \ln 48 - \ln 3$, which simplifies to $8 \ln(16 \cdot 3) - \ln 3 = 32 \ln 2 + 7 \ln 3$, identifying $\alpha=32$ and $\beta=7$ with a sum of 39.

Computing $\int \frac{2 x^{2+}5 x+9}{\sqrt{x^{2+}x+1}} dx$ requires expressing the quadratic numerator in terms of the radicand $x^{2+}x+1$ and its derivative $2x+1$, yielding $2(x^{2+}x+1) + \frac{3}{2}(2x+1) + \frac{11}{2}$. Splitting this over the denominator results in three integrals: $2 \int \sqrt{x^{2+}x+1} dx$, $\frac{3}{2} \int (2x+1)(x^{2+}x+1)^{-1/2} dx$, and $\frac{11}{2} \int (x^{2+}x+1)^{-1/2} dx$. The first integral applies the standard form for $\int \sqrt{u^{2+}a^2} du$, which results in $(x+\frac{1}{2})\sqrt{x^{2+}x+1} + \frac{3}{4} \ln |x+\frac{1}{2}+\sqrt{x^{2+}x+1}|$. The second integral simplifies to $3\sqrt{x^{2+}x+1}$ by power rule integration, and the third integral yields $\frac{11}{2} \ln |x+\frac{1}{2}+\sqrt{x^{2+}x+1}|$. Combining the algebraic terms produces $(x + \frac{1}{2} + 3)\sqrt{x^{2+}x+1}$, which simplifies to $x\sqrt{x^{2+}x+1} + \frac{7}{2}\sqrt{x^{2+}x+1}$, identifying $\alpha = \frac{7}{2}$. Combining the logarithmic terms gives $\beta = \frac{3}{4} + \frac{11}{2} = \frac{25}{4}$. Calculating $\alpha + 2\beta$ leads to $\frac{7}{2} + 2(\frac{25}{4}) = \frac{7}{2} + \frac{25}{2} = 16$.

The core principle behind this integration involves simplifying the integrand to reveal a form suitable for substitution, specifically by factoring out an appropriate power of $x$ from the second term. To begin, examine the expression inside the exponent: $\frac{3}{x^{24}} + \frac{1}{x^{26}}$. By factoring out $\frac{1}{x^{23}}$, the expression becomes $\frac{1}{x^{23}}(\frac{3}{x} + \frac{1}{x^3})$. The integral can then be rewritten as follows:
$\int \left( \frac{1}{x} + \frac{1}{x^3} \right) \left( \frac{1}{x^{23}} \left( \frac{3}{x} + \frac{1}{x^3} \right) \right)^{\frac{1}{23}} dx$
Applying the power of the exponent to $\frac{1}{x^{23}}$, the term simplifies to $\frac{1}{x}$, which, when multiplied by $(\frac{1}{x} + \frac{1}{x^3})$, yields $(\frac{1}{x^2} + \frac{1}{x^4})$. This transforms the integral into a much simpler form:
$\int \left( \frac{1}{x^2} + \frac{1}{x^4} \right) \left( \frac{3}{x} + \frac{1}{x^3} \right)^{\frac{1}{23}} dx$
By substituting $u = \frac{3}{x} + \frac{1}{x^3}$, the differential becomes $du = (-3x^{-2} - 3x^{-4}) dx$, or $du = -3(\frac{1}{x^2} + \frac{1}{x^4}) dx$. Therefore, the term $(\frac{1}{x^2} + \frac{1}{x^4}) dx$ is equivalent to $-\frac{1}{3} du$. The integral is now solvable via the standard power rule:
$-\frac{1}{3} \int u^{\frac{1}{23}} du = -\frac{1}{3} \cdot \frac{u^{\frac{1}{23} + 1}}{\frac{1}{23} + 1} + C = -\frac{1}{3} \cdot \frac{u^{\frac{24}{23}}}{\frac{24}{23}} + C = -\frac{23}{72} \left( \frac{3}{x} + \frac{1}{x^3} \right)^{\frac{24}{23}} + C$
Comparing this to the given expression $-\frac{\alpha}{3(\alpha+1)} (3 x^\beta + x^\gamma)^{\frac{\alpha+1}{\alpha}} + C$, we identify $\alpha = 23$, $\beta = -1$, and $\gamma = -3$. The requested sum is therefore $\alpha + \beta + \gamma = 23 - 1 - 3 = 19$.

The integrand simplifies by noting that $(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x) = 1$, which implies $(\sqrt{1+x^2}-x)^{-9} = (\sqrt{1+x^2}+x)^9$. Multiplying the terms, the integral becomes $\int (\sqrt{1+x^2}+x)^{19} dx$. Letting $y = \sqrt{1+x^2}+x$, the integral of $y^n$ is given by $\int y^n dx = \frac{y^n (n\sqrt{1+x^2}-x)}{n^{2-}1} + C$ for $n \neq 1$. Matching this with the provided form $\frac{1}{m} (y^n (n\sqrt{1+x^2}-x)) + C$, we identify $n=19$ and $m = n^{2-}1 = 19^{2-}1 = 360$. Thus, $m+n = 360+19 = 379$.

Determining the value of $f(1)$ begins by using the substitution $t^2 = 3-x^2$ to simplify the radicand, which also requires transforming the differential $x \, dx = -t \, dt$. This transformation turns the integral into $\int (3-t^2) \cdot t \cdot (-t) \, dt$, or $\int (t^4 - 3t^2) \, dt$, which integrates to $\frac{t^5}{5} - t^3 + c$. Substituting the original variable back into the expression yields $f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + c$. Since $5f(\sqrt{2}) = -4$, the value of the function at $x=\sqrt{2}$ is $-\frac{4}{5}$, so substituting this into the equation results in $\frac{1}{5} - 1 + c = -\frac{4}{5}$, which confirms that the constant $c = 0$. Finally, evaluating the function at $x=1$ gives $\frac{(3-1)^{5/2}}{5} - (3-1)^{3/2}$, which simplifies to $\frac{2^{5/2}}{5} - 2^{3/2}$. Factoring out $2\sqrt{2}$ from this expression leads to $2\sqrt{2} \left( \frac{2}{5} - 1 \right)$, resulting in a final value of $-\frac{6\sqrt{2}}{5}$.

Consolidating the integrand relies on grouping the terms under a shared fractional power to reveal a standard substitution. Multiplying and dividing by $(x-11)^{\frac{15}{13}}$ transforms the integral into

$I(x) = \int \frac{dx}{(x-11)^2 \left( \frac{x+15}{x-11} \right)^{\frac{15}{13}}}$

Defining $t = \frac{x+15}{x-11}$ introduces the differential $dt = -\frac{26}{(x-11)^2} dx$, which allows the integral to be solved as $-\frac{1}{26} \int t^{-\frac{15}{13}} dt$, ultimately yielding $I(x) = \frac{1}{4} \left( \frac{x-11}{x+15} \right)^{\frac{2}{13}}$.

Evaluating this expression at the boundaries $x=37$ and $x=24$ provides the required difference. At $x=37$, the ratio becomes $\frac{37-11}{37+15} = \frac{26}{52} = \frac{1}{2}$, and at $x=24$, the ratio is $\frac{24-11}{24+15} = \frac{13}{39} = \frac{1}{3}$. Subtracting these values provides $I(37) - I(24) = \frac{1}{4} \left( \left(\frac{1}{2}\right)^{\frac{2}{13}} - \left(\frac{1}{3}\right)^{\frac{2}{13}} \right)$. Since $\left(\frac{1}{2}\right)^{\frac{2}{13}} = \left(\frac{1}{4}\right)^{\frac{1}{13}}$ and $\left(\frac{1}{3}\right)^{\frac{2}{13}} = \left(\frac{1}{9}\right)^{\frac{1}{13}}$, identifying the constants yields $b=4$ and $c=9$. Computing $3(b+c)$ results in $3(4+9)$, which equals $39$.

The integration of $x^3 \sin x$ by parts involves repeatedly differentiating $x^3$ and integrating $\sin x$, which produces the expression
$g(x) = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x$
By definition, the derivative $g'(x)$ must equal the original integrand, $x^3 \sin x$. When substituting the value $x = \frac{\pi}{2}$ into these expressions, the terms involving $\cos \frac{\pi}{2}$ vanish while $\sin \frac{\pi}{2}$ becomes 1, simplifying the values to $g\left(\frac{\pi}{2}\right) = \frac{3 \pi^2}{4} - 6$ and $g'\left(\frac{\pi}{2}\right) = \frac{\pi^3}{8}$.

Substituting these results into the specified expression yields
$8 \left( g\left(\frac{\pi}{2}\right) + g'\left(\frac{\pi}{2}\right) \right) = 8 \left( \frac{3 \pi^2}{4} - 6 + \frac{\pi^3}{8} \right) = \pi^3 + 6 \pi^2 - 48$
Equating this result to the standard form $\alpha \pi^3 + \beta \pi^2 + \gamma$ identifies the constants as $\alpha = 1$, $\beta = 6$, and $\gamma = -48$. Consequently, the value of $\alpha + \beta - \gamma$ is $1 + 6 - (-48)$, which equals 55.

When dealing with integrands containing fractional roots, a variable substitution to eliminate the radical often transforms the expression into a manageable rational function. Substituting $t = x^{1/4}$ implies $dx = 4t^3 \, dt$, which simplifies the original integral to $4 \int \frac{t^2}{t+1} \, dt$. By rewriting the integrand as $t - 1 + \frac{1}{t+1}$, the integration yields the general solution:

$f(x) = 4 \left( \frac{x^{1/2}}{2} - x^{1/4} + \ln(x^{1/4} + 1) \right) + C$

Using the initial condition $f(0) = -6$, we find that $C = -6$. Finally, evaluating the expression at $x = 1$ leads to $4 \left( \frac{1}{2} - 1 + \ln(2) \right) - 6$, which simplifies to $4(\log_e 2 - 2)$.

The integrand follows the standard form $\int e^x (f(x) + f'(x)) \, dx$, which evaluates to $e^x f(x) + C$. By defining $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$, we can verify the other terms in the integrand by differentiating $f(x)$ using the product rule. Differentiating this function yields $f'(x) = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$, which perfectly matches the remaining portion of the original expression. Consequently, the integral simplifies to $g(x) + C$, where $g(x) = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.

Evaluating $g\left( \frac{1}{2} \right)$ involves substituting $x = \frac{1}{2}$ into this expression. Since $e^{1/2} = \sqrt{e}$, $\sin^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{6}$, and the denominator $\sqrt{1 - (1/2)^2} = \sqrt{3/4} = \frac{\sqrt{3}}{2}$, the value becomes $g\left( \frac{1}{2} \right) = \sqrt{e} \cdot \frac{(1/2)(\pi/6)}{\sqrt{3}/2}$. Simplifying this fraction, we have $\sqrt{e} \cdot \frac{\pi/12}{\sqrt{3}/2} = \sqrt{e} \cdot \frac{\pi}{6\sqrt{3}}$, which rearranges to $\frac{\pi}{6} \sqrt{\frac{e}{3}}$.

The core strategy for this integral involves transforming the complex trigonometric expression into a rational function of $\sin x$ using standard identities, which sets the stage for a direct substitution. By converting all terms into sine and cosine, the integrand simplifies to $\frac{(1+\sin^2 x)\cos x}{1+\sin^4 x}$. Applying the substitution $t = \sin x$ allows us to rewrite the integral as $\int \frac{1+t^2}{1+t^4} \, dt$.

To solve this resulting integral, divide both the numerator and the denominator by $t^2$, which transforms the expression into $\int \frac{1+1/t^2}{t^{2+}1/t^2} \, dt$. By rewriting the denominator as $(t - 1/t)^2 + 2$, the integral becomes a standard form that yields $\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t - 1/t}{\sqrt{2}} \right) + C$.

The limit condition $\lim_{x \rightarrow \frac{\pi}{2}^-} y(x) = 0$ requires that when $t = 1$, the function value is zero, which determines that the integration constant $C$ must be $0$. Consequently, evaluating the function at $x = \frac{\pi}{4}$, where $t = \frac{1}{\sqrt{2}}$, involves calculating $\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{1/\sqrt{2} - \sqrt{2}}{\sqrt{2}} \right)$. This simplifies directly to $\frac{1}{\sqrt{2}} \tan^{-1} \left(-\frac{1}{2}\right)$.

Rewriting the integrand $\frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}}$ as $\frac{1}{(x-1)^{4/5}(x+3)^{6/5}}$ allows for a strategic substitution. Factoring out $(x+3)^2$ from the denominator transforms the expression into $\frac{1}{\left( \frac{x-1}{x+3} \right)^{4/5} (x+3)^2} dx$. By choosing $t = \frac{x-1}{x+3}$, the differential becomes $dt = \frac{4}{(x+3)^2} dx$, which simplifies the integral to $\frac{1}{4} \int t^{-4/5} dt$.