Indefinite Integrals – JEE Main MathematicsPractice Questions & PYQs

Simplifying the integrand begins by factoring $\sin^n \theta$ out of the term raised to the power of $1/n$. This transforms the numerator into $\sin \theta (1 - \frac{1}{\sin^{n-1} \theta})^{1/n}$, allowing the factor $\sin \theta$ to reduce the power of the denominator from $\sin^{n+1} \theta$ to $\sin^n \theta$. The integral is then effectively expressed as $\int (1 - \frac{1}{\sin^{n-1} \theta})^{1/n} \frac{\cos \theta}{\sin^n \theta} d\theta$.

Choosing $t = 1 - \frac{1}{\sin^{n-1} \theta}$ provides a straightforward substitution because differentiating this with respect to $\theta$ yields $dt = (n-1) \frac{\cos \theta}{\sin^n \theta} d\theta$. This differential corresponds perfectly with the remaining terms in the integral, allowing the expression to be rewritten as $\frac{1}{n-1} \int t^{1/n} dt$. Applying the standard power rule for integration results in $\frac{1}{n-1} \cdot \frac{t^{(n+1)/n}}{(n+1)/n} + C$, which simplifies to the final result of $\frac{n}{n^{2-}1} (1 - \frac{1}{\sin^{n-1} \theta})^{\frac{n+1}{n}} + C$.

Integration by substitution provides the most direct path to solving this integral, starting by simplifying the exponent. Setting $u = x^{2}$ makes the differential $du = 2x dx$, which transforms the expression $x^{5} e^{-x^{2}} dx$ into $\frac{1}{2} u^{2} e^{-u} du$. Applying integration by parts, where $u^{2}$ is treated as the algebraic part and $e^{-u}$ is integrated repeatedly, yields the result $-\frac{1}{2} e^{-u} (u^{2} + 2u + 2) + c$.

Substituting $x^{2}$ back for $u$ allows the integral to be expressed as $-\frac{1}{2} e^{-x^{2}} (x^{4} + 2x^{2} + 2) + c$. Comparing this form to $g(x) e^{-x^{2}} + c$ isolates the function $g(x) = -\frac{1}{2} (x^{4} + 2x^{2} + 2)$. Evaluating this function at $x = -1$ leads to $g(-1) = -\frac{1}{2} ((-1)^{4} + 2(-1)^{2} + 2)$, which simplifies to $-\frac{1}{2} (1 + 2 + 2)$, resulting in $-\frac{5}{2}$.

To resolve integrals containing a square root of a linear expression, substitution is the most efficient method. By setting $t = \sqrt{2x-1}$, we find that $t^2 = 2x - 1$, which leads to $x = \frac{t^{2+}1}{2}$ and subsequently yields the differential $dx = t \, dt$. Substituting these expressions into the original integral transforms it into $\int \frac{\frac{t^{2+}1}{2}+1}{t} \cdot t \, dt$. Since the $t$ terms cancel out, the integral simplifies to the more manageable form $\int \frac{t^{2+}3}{2} \, dt$.

Integrating this expression with respect to $t$ results in $\frac{1}{2}(\frac{t^3}{3} + 3t) + C$, which can be simplified by factoring out $t$ to obtain $\frac{t}{6}(t^{2+}9) + C$. Replacing $t$ back with $\sqrt{2x-1}$ and substituting the definition $t^2 = 2x-1$ into the parentheses gives $\frac{\sqrt{2x-1}}{6}(2x-1+9) + C$. This simplifies to $\frac{\sqrt{2x-1}}{6}(2x+8) + C$, and by factoring a 2 from the binomial, the expression becomes $\frac{\sqrt{2x-1}}{3}(x+4) + C$. Comparing this result to the required form $f(x) \sqrt{2x-1} + C$ identifies $f(x)$ as $\frac{1}{3}(x+4)$.

Simplifying the integrand through trigonometric identities allows us to convert the quotient of sine functions into a straightforward sum of terms. By expressing the ratio $\frac{\sin\frac{5x}{2}}{\sin\frac{x}{2}}$ as $1 + 2\cos x + 2\cos 2x$, the problem transforms from a complex fraction into a set of basic trigonometric integrals. Each part can then be evaluated individually using standard calculus rules.

Integrating $1$ with respect to $x$ provides $x$, while the integral of $2\cos x$ produces $2\sin x$. Finally, the integral of $2\cos 2x$ contributes $\sin 2x$ to the result. Summing these individual integrals yields $x + 2 \sin x + \sin 2x + c$.

The problem relies on the fundamental relationship between an integral and its derivative, specifically recognizing that the given expression follows the product rule form \frac{d}{dx} \[e^{\sec x} f(x)]$ = e^{\sec x} (\sec x \tan x) f(x) + e^{\sec x} f'(x)$. By equating this derivative to the integrand on the left-hand side of the original equation, the term$e^{\sec x} (\sec x \tan x) f(x)$cancels out, which simplifies the equation to$f'(x) = \sec x \tan x + \sec^2 x$. Integrating this expression with respect to$x$yields$f(x) = \int (\sec x \tan x + \sec^2 x) dx$, resulting in$f(x) = \sec x + \tan x + c$. Aligning this result with the provided options, the constant$c$is represented by$\frac{1}{2}$, confirming that$\sec x + \tan x + \frac{1}{2}$is the expression for$f(x)$.

Solving the integral $\int \cos(\log_e x) \, dx$ requires the technique of integration by parts, assigning $u = \cos(\log_e x)$ and $dv = dx$ to yield $du = -\frac{\sin(\log_e x)}{x} \, dx$ and $v = x$. Applying the formula $\int u \, dv = uv - \int v \, du$ creates the equation $I = x \cos(\log_e x) - \int x \cdot \left(-\frac{\sin(\log_e x)}{x}\right) \, dx = x \cos(\log_e x) + \int \sin(\log_e x) \, dx$ Performing integration by parts on the remaining integral $\int \sin(\log_e x) \, dx$ with $u = \sin(\log_e x)$ and $dv = dx$ produces $du = \frac{\cos(\log_e x)}{x} \, dx$ and $v = x$. Substituting this back yields $I = x \cos(\log_e x) + \left(x \sin(\log_e x) - \int x \cdot \frac{\cos(\log_e x)}{x} \, dx\right)$ Because the integral term on the right side is identical to the original integral $I$, the expression simplifies to $I = x \cos(\log_e x) + x \sin(\log_e x) - I$. Adding $I$ to both sides results \in $2I = x (\cos(\log_e x) + \sin(\log_e x))$, and dividing by two gives the final value of $\frac{x}{2} [\cos(\log_e x) + \sin(\log_e x)] + C$.

The core of this problem lies in simplifying the trigonometric expression by converting the tangent functions into ratios of sine and cosine, which transforms the integrand into $\frac{\sin(x+\alpha)}{\sin(x-\alpha)}$. By decomposing the numerator as $\sin((x-\alpha) + 2\alpha)$, we apply the sine addition formula to rewrite the expression as $\cos(2\alpha) + \cot(x-\alpha)\sin(2\alpha)$. Integrating this with respect to $x$ yields $x \cos(2\alpha) + \sin(2\alpha) \log_e |\sin(x-\alpha)| + C$. Because any constant can be absorbed into the integration constant $C$, the term $x \cos(2\alpha)$ is equivalent to $(x-\alpha) \cos(2\alpha)$ plus an additional constant term. This adjustment directly yields the required form, identifying $A(x)$ as $x - \alpha$ and $B(x)$ as $\log_e |\sin(x-\alpha)|$.

Expressing trigonometric integrals in terms of tangent and secant functions often reveals an immediate substitution that simplifies the expression. The integral $\int \sec^{\frac{2}{3}} x \csc^{\frac{4}{3}} x \, dx$ can be rewritten in terms of sine and cosine as
$\int \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$
Multiplying the integrand by $\frac{\cos^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x}$ allows for the grouping of sine and cosine into a tangent term, since $\frac{\sin^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x} = \tan^{\frac{4}{3}} x$:
$\int \frac{\cos^{\frac{4}{3}} x}{\cos^2 x \sin^{\frac{4}{3}} x} \, dx = \int \frac{1}{\cos^2 x (\tan x)^{\frac{4}{3}}} \, dx = \int \sec^2 x (\tan x)^{-\frac{4}{3}} \, dx$
Applying the substitution $t = \tan x$ gives $dt = \sec^2 x \, dx$, which transforms the integral into $\int t^{-\frac{4}{3}} \, dt$. Performing the integration via the power rule results in $-3 t^{-\frac{1}{3}} + C$, which simplifies to $-3 \tan^{-\frac{1}{3}} x + C$.

Simplifying the trigonometric expression within the square root involves applying the double-angle formula $\sin 2A = 2\sin A \cos A$ to reduce the complexity of the integrand. Substituting $\theta = \frac{x^{2-}1}{2}$, where $x \, dx = d\theta$, transforms the expression into $\int \sqrt{\frac{2\sin 2\theta - \sin 4\theta}{2\sin 2\theta + \sin 4\theta}} \, d\theta$. Expanding $\sin 4\theta$ as $2\sin 2\theta \cos 2\theta$ allows us to factor and cancel $2\sin 2\theta$ from the numerator and denominator, resulting in $\int \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} \, d\theta$. Since $1 - \cos 2\theta = 2\sin^2 \theta$ and $1 + \cos 2\theta = 2\cos^2 \theta$, the integral simplifies to $\int \tan \theta \, d\theta$, which evaluates to $\ln \left| \sec\left( \frac{x^{2-}1}{2} \right) \right| + C$.

The technique of trigonometric substitution effectively simplifies integrals containing a squared quadratic expression, such as $\frac{1}{(t^{2+}a^2)^2}$, by setting $t = a\tan\theta$ to reduce the expression to a power of the cosine function. Completing the square for the integral of $\frac{1}{(x^{2-}2x+10)^2}$ involves substituting $t=x-1$, which transforms the integrand into $\int \frac{dt}{(t^{2+}9)^2}$. Using $t=3\tan\theta$ leads to the integral $\frac{1}{27} \int \cos^2\theta d\theta$, which evaluates to $\frac{1}{54}\theta + \frac{1}{108}\sin(2\theta) + C$. Expressing $\sin(2\theta)$ as $\frac{2\tan\theta}{1+\tan^2\theta}$ and substituting $\tan\theta = \frac{t}{3}$ yields $\frac{6t}{t^{2+}9}$, simplifying the full expression to $\frac{1}{54}\tan^{-1}(\frac{t}{3}) + \frac{1}{18}(\frac{t}{t^{2+}9}) + C$. Factoring out $\frac{1}{54}$ to align with the required form results in $\frac{1}{54}(\tan^{-1}(\frac{x-1}{3}) + \frac{3(x-1)}{x^{2-}2x+10}) + C$, identifying $A = \frac{1}{54}$ and $f(x) = 3(x-1)$.

Integrating rational functions is often simplified by adjusting the numerator to align with factors in the denominator. Factoring $x^4 + x$ as $x(x^3 + 1)$ allows $2x^3 - 1$ to be rewritten as $2(x^3 + 1) - 3$, splitting the integral into:
$\int \frac{2}{x} \, dx - \int \frac{3}{x(x^3 + 1)} \, dx$

Multiplying the second integrand by $\frac{x^2}{x^2}$ introduces $d(x^3)$ in the numerator, transforming the expression into $\int \frac{d(x^3)}{x^3(x^3 + 1)}$. Using partial fractions, the integrand $\frac{1}{x^3(x^3 + 1)}$ becomes $\frac{1}{x^3} - \frac{1}{x^3 + 1}$, which integrates to $\log |x^3| - \log |x^3 + 1|$.

Combining the components yields $2\log |x| - \log |x^3| + \log |x^3 + 1| + C$. Using the identity $2\log |x| = \log |x^2|$, the expression simplifies to $\log |x^2| - \log |x^3| + \log |x^3 + 1| + C$, which ultimately reduces to:
$\log \left| \frac{x^3 + 1}{x} \right| + C$

The integral $\int \frac{px+q}{\sqrt{ax^{2+}bx+c}} dx$ is evaluated by expressing the numerator as a linear combination of the derivative of the quadratic expression and a constant, specifically $2x+5 = -1(-2x-6) - 1$. Substituting this into the integral gives $\int \frac{-(-2x-6)-1}{\sqrt{7-6x-x^2}} dx$, which separates into $-\int \frac{-2x-6}{\sqrt{7-6x-x^2}} dx - \int \frac{1}{\sqrt{16-(x+3)^2}} dx$. Evaluating these terms using standard integral forms yields $-2\sqrt{7-6x-x^2} - \sin^{-1} \left( \frac{x+3}{4} \right) + C$. Comparing this result to the given expression $A\sqrt{7-6x-x^2}+B \sin^{-1} \left( \frac{x+3}{4} \right) + C$ identifies the constants as $A = -2$ and $B = -1$.

The substitution $u = \tan x$ results in $du = (1+u^2) \, dx$, transforming the integral into $\int \frac{u}{(1+u+u^2)(1+u^2)} \, du$. Applying partial fraction decomposition to the integrand yields $\frac{1}{1+u^2} - \frac{1}{u^{2+}u+1}$.

Integrating these terms individually results in $\tan^{-1} u - \int \frac{1}{(u+1/2)^2 + 3/4} \, du$, which evaluates to $\tan^{-1} u - \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2u+1}{\sqrt{3}}\right) + C$. Substituting $u = \tan x$ back into this expression provides $x - \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2\tan x + 1}{\sqrt{3}}\right) + C$, allowing the values $K=2$ and $A=3$ to be identified by direct comparison with the given form.

Substitution of $t = \tan x$ transforms the integral into the algebraic form $\int \frac{t^2 dt}{(t^{3+}1)^2}$. Applying the substitution $u = t^3 + 1$, where $du = 3t^2 dt$, simplifies the integral to $\frac{1}{3} \int u^{-2} du$. Evaluating this power integral yields $\frac{-1}{3u} + C$, which, upon substituting $u = 1+\tan^3 x$, results in $\frac{-1}{3(1+\tan^3 x)} + C$.

The substitution method allows for the determination of an explicit form for a function given a composite argument by setting $t = \frac{x-4}{x+2}$ and expressing $x$ in terms of $t$. Solving for $x$ yields $t(x+2) = x-4$, which simplifies to $tx + 2t = x-4$, leading to $x(t-1) = -2t-4$, and finally $x = \frac{2t+4}{1-t}$.

Substituting this expression for $x$ into the given function $f(t) = 2x+1$ results in $f(t) = 2\left(\frac{2t+4}{1-t}\right) + 1 = \frac{4t+8+1-t}{1-t} = \frac{3t+9}{1-t}$. Replacing $t$ with $x$, the function is defined as $f(x) = \frac{3x+9}{1-x}$.

The integral $\int f(x) \, dx = \int \frac{3x+9}{1-x} \, dx$ is evaluated by expressing the numerator to match the denominator, $\int \frac{-3(1-x) + 12}{1-x} \, dx = \int \left(-3 + \frac{12}{1-x}\right) \, dx$. Integrating these terms individually yields $-3x + 12 \frac{\log_e |1-x|}{-1} + C$, which simplifies to $-3x - 12 \log_e |1-x| + C$.

To evaluate the sum of the integrals $I_4 + I_6$, we first combine the terms under a single integral sign:
$\int (\tan^4 x + \tan^6 x) \, dx$
Factoring $\tan^4 x$ from the integrand gives $\tan^4 x(1 + \tan^2 x)$, which simplifies to $\tan^4 x \sec^2 x$ by applying the trigonometric identity $\sec^2 x = 1 + \tan^2 x$. We can then solve the integral by using the substitution $u = \tan x$, which implies $du = \sec^2 x \, dx$, transforming the expression into $\int u^4 \, du$. Performing the integration results in $\frac{1}{5} u^5 + C$, which reverts to $\frac{1}{5} \tan^5 x + C$. Matching this result with the given expression $a \tan^5 x + b x^5 + C$ allows us to identify the coefficients as $a = \frac{1}{5}$ and $b = 0$, yielding the ordered pair $\left(\frac{1}{5}, 0\right)$.

The expression $1+2\cot x(\csc x+\cot x)$ simplifies by applying the fundamental trigonometric identity $\csc^2 x - \cot^2 x = 1$. Substituting this value into the expression transforms it into $\csc^2 x - \cot^2 x + 2\csc x \cot x + 2\cot^2 x$, which combines to form $\csc^2 x + 2\csc x \cot x + \cot^2 x$, a perfect square equivalent to $(\csc x + \cot x)^2$. Taking the square root of this square leaves the integrand as $\csc x + \cot x$. By converting this to $\frac{1+\cos x}{\sin x}$, the integrand simplifies to $\cot(x/2)$ using standard half-angle formulas. Integrating $\cot(x/2)$ with respect to $x$ yields $2\log(\sin(x/2)) + c$.

Substitution of $t = \frac{3x-4}{3x+4}$ allows solving for $x$ as $x = \frac{4(1+t)}{3(1-t)}$, which leads to $f(t) = \frac{4(1+t)}{3(1-t)} + 2 = \frac{4+4t+6-6t}{3(1-t)} = \frac{10-2t}{3(1-t)}$. Rewriting $f(x) = \frac{10-2x}{3(1-x)}$ as $\frac{2(1-x)+8}{3(1-x)} = \frac{2}{3} + \frac{8}{3(1-x)}$ enables direct integration.

The integral $\int (\frac{2}{3} + \frac{8}{3(1-x)})dx$ evaluates to $\frac{2}{3}x - \frac{8}{3}\ln|1-x| + C$. Comparing this expression to $A\log|1-x|+Bx+C$ identifies $A = -\frac{8}{3}$ and $B = \frac{2}{3}$, yielding the ordered pair $\left(-\frac{8}{3}, \frac{2}{3}\right)$.

Rewriting the denominator as $(1+\sqrt{x})\sqrt{x}\sqrt{1-x}$ suggests using the trigonometric substitution $x = \cos^2 \theta$, which implies $dx = -2 \cos \theta \sin \theta \, d\theta$. Substituting these into the integral allows the $\cos \theta$ and $\sin \theta$ terms to cancel, simplifying the integrand to $\frac{-2}{1+\cos \theta} \, d\theta$. Applying the half-angle identity $1+\cos \theta = 2 \cos^2(\theta/2)$ reduces the integral to $-\int \sec^2(\theta/2) \, d\theta$, which evaluates to $-2 \tan(\theta/2) + C$. Given that $\tan^2(\theta/2) = \frac{1-\cos \theta}{1+\cos \theta}$ and $\cos \theta = \sqrt{x}$, we find $\tan(\theta/2) = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$, leading to the final result $-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$.