Indefinite Integrals – JEE Main MathematicsPractice Questions & PYQs

To solve the integral, rewrite the integrand by using the identity $\sin 2x = 2 \sin x \cos x$. The denominator simplifies to $\cos^3 x \sqrt{4 \sin x \cos x}$, which evaluates to $2 \cos^{7/2} x \sin^{1/2} x$. By expressing the integrand in terms of $\tan x$, we obtain:

$\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = \frac{1}{2} \int (1 + \tan^2 x) \sec^2 x (\tan x)^{-1/2} dx$

Substitute $t = \tan x$, where $dt = \sec^2 x dx$, to transform the integral into $\frac{1}{2} \int (t^{-1/2} + t^{3/2}) dt$. This results in $t^{1/2} + \frac{1}{5} t^{5/2} + k$. Comparing this to the given expression $(\tan x)^A + C (\tan x)^B + k$, we identify $A = 1/2$, $C = 1/5$, and $B = 5/2$. The sum of these values is $\frac{1}{2} + \frac{1}{5} + \frac{5}{2} = \frac{16}{5}$.

Factoring $x^4$ out of the term $(x^{4+}1)^{3/4}$ transforms the integral into a manageable form by utilizing the substitution method. By rewriting the denominator, the expression becomes
$\int \frac{dx}{x^2 (x^4)^{3/4} (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$
Applying the substitution $t = 1 + \frac{1}{x^4}$ yields the differential $dt = -\frac{4}{x^5} dx$, which simplifies the integral to $-\frac{1}{4} \int t^{-3/4} dt$. Evaluating this integral results in $-t^{1/4} + c$, and substituting the definition of $t$ back into the result yields $-\left(\frac{x^{4+}1}{x^4}\right)^{1/4} + c$.

When solving integrals that feature exponential expressions like $e^{x+\frac{1}{x}}$, a helpful strategy is to manipulate the integrand to isolate the derivative of the exponent, which in this case is $1 - \frac{1}{x^2}$. By splitting the original integrand $\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}}$ into two components, we can rewrite the expression as $\int e^{x+\frac{1}{x}} dx + \int x \left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} dx$.

This rearrangement allows for the application of integration by parts specifically to the second term. By choosing $u = x$ and $dv = \left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} dx$, we find that $du = dx$ and $v = e^{x+\frac{1}{x}}$. Applying the integration by parts formula yields $x e^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}} dx$ for the second portion. When this result is combined with the first integral we set aside, $\int e^{x+\frac{1}{x}} dx$, the two integral terms cancel one another, simplifying the final expression to $x e^{x+\frac{1}{x}} + c$.

Substitution effectively simplifies integrals containing composite functions by changing the variable, which makes it easier to apply the technique of integration by parts. Given $\int f(x) dx = \Psi(x)$, applying the change of variable $t = x^3$ leads to the differential $dt = 3x^2 dx$, or $x^2 dx = \frac{1}{3} dt$. Consequently, the integral $\int x^5 f(x^3) dx$ can be rewritten as $\int x^3 f(x^3) \cdot x^2 dx$, which transforms into $\int t f(t) \cdot \frac{1}{3} dt$, or $\frac{1}{3} \int t f(t) dt$.

Integrating $\frac{1}{3} \int t f(t) dt$ by parts requires assigning $u = t$ and $dv = f(t) dt$, which gives $du = dt$ and $v = \Psi(t)$. Using the product rule $\int u dv = uv - \int v du$ results in $\frac{1}{3} [t \Psi(t) - \int 1 \cdot \Psi(t) dt]$. Substituting $t = x^3$ and converting the integral to the variable $x$ by replacing $dt$ with $3x^2 dx$ produces $\frac{1}{3} x^3 \Psi(x^3) - \frac{1}{3} \int \Psi(x^3) \cdot 3x^2 dx$, which simplifies to $\frac{1}{3} x^3 \Psi(x^3) - \int x^2 \Psi(x^3) dx + C$.

The integral $\int \frac{5 \tan x}{\tan x - 2} \, dx$ is simplified by rewriting $\tan x$ as $\frac{\sin x}{\cos x}$, which transforms the integrand into the form:
$\int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx$
To evaluate this expression, represent the numerator as a linear combination of the denominator, $\sin x - 2 \cos x$, and its derivative, $\cos x + 2 \sin x$. Specifically, the numerator $5 \sin x$ can be split into $1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)$, allowing the integral to be decomposed as $\int 1 \, dx + 2 \int \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} \, dx$. The first part evaluates directly to $x$, while the second part takes the form $\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C$. This results in $x + 2 \ln | \sin x - 2 \cos x | + k$, and by comparing this result to the given expression $x + a \ln | \sin x - 2 \cos x | + k$, we identify that $a = 2$.

The integrand can be simplified by applying the substitution $t = x - \frac{\pi}{4}$, which implies $dx = dt$ and transforms $\sin x$ into $\sin \left( t + \frac{\pi}{4} \right)$. Using the trigonometric identity for the sine of a sum, $\sin \left( t + \frac{\pi}{4} \right)$ becomes $\frac{1}{\sqrt{2}} (\sin t + \cos t)$, so the total expression simplifies significantly when multiplied by $\sqrt{2}$. This reduction yields the integral $\int (1 + \cot t) \, dt$, which evaluates directly to $t + \log |\sin t| + c_1$. Replacing $t$ with the original variable $x - \frac{\pi}{4}$ and incorporating the resulting constant $-\frac{\pi}{4}$ into a new arbitrary constant $c$ gives the final form $x + \log \left| \sin \left( x - \frac{\pi}{4} \right) \right| + c$.

The denominator $\cos x + \sqrt{3} \sin x$ is equivalent to $2(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x)$, which simplifies using the sine addition identity to $2\sin(x + \frac{\pi}{6})$. This conversion transforms the integral into $\frac{1}{2}\int \csc(x + \frac{\pi}{6}) dx$.

Since the standard integral of $\csc \theta$ is $\log\tan(\frac{\theta}{2}) + C$, applying this to the transformed expression requires dividing the argument $(x + \frac{\pi}{6})$ by $2$. This yields the result $\frac{1}{2}\log\tan(\frac{x}{2} + \frac{\pi}{12}) + C$.

Expanding the squared expression in the integrand results in $\frac{(\log x)^2 - 2 \log x + 1}{[1+(\log x)^2]^2}$, which simplifies into the difference of two terms: $\frac{1}{1+(\log x)^2} - \frac{2 \log x}{[1+(\log x)^2]^2}$. By applying the substitution $t = \log x$, we have $dx = e^t dt$, which transforms the integral into \int e^t \left[ \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right]\ dt$. Identifying$f(t) = \frac{1}{1+t^2}$, we observe that its derivative is$f'(t) = -\frac{2t}{(1+t^2)^2}$. Since the integral matches the standard form$\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$, the expression evaluates to$\frac{e^t}{1+t^2} + C$. Replacing$e^t$with$x$and$t$with$\log x$gives the final result$\frac{x}{1+(\log x)^2} + C$.

Rewriting the numerator as $\sin((x-\alpha) + \alpha)$ effectively aligns the argument with the denominator. Expanding this using the sine addition rule yields $\sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha$. Dividing through by $\sin(x-\alpha)$ simplifies the integrand to $\cos\alpha + \sin\alpha \cot(x-\alpha)$. Integrating these components produces $x \cos\alpha + \sin\alpha \ln \sin(x-\alpha) + C$. Comparison with the form $Ax + B \log \sin (x-\alpha) + C$ reveals that $A = \cos\alpha$ and $B = \sin\alpha$.

Evaluating the integral $\int \frac{dx}{\cos x - \sin x}$ begins by simplifying the expression in the denominator. Factoring out $\sqrt{2}$ yields $\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \right)$, which transforms into $\sqrt{2} \cos \left( x + \frac{\pi}{4} \right)$ using the cosine angle addition formula, where $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

The integral then becomes $\frac{1}{\sqrt{2}} \int \sec \left( x + \frac{\pi}{4} \right) \, dx$. Utilizing the standard integration result $\int \sec \theta \, d\theta = \ln \left| \tan \left( \frac{\theta}{2} + \frac{\pi}{4} \right) \right| + C$ with $\theta = x + \frac{\pi}{4}$, we substitute the value of $\theta$ into the formula. This results in $\frac{1}{\sqrt{2}} \ln \left| \tan \left( \frac{x}{2} + \frac{\pi}{8} + \frac{\pi}{4} \right) \right| + C$, which simplifies to the final expression $\frac{1}{\sqrt{2}} \ln \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + C$.