JEE Indefinite Integration PYQs with Video Solutions PDF

Let $$t = 2x - 1$$, so $$dt = 2\,dx$$. Then $$4x^2 - 4x + 6 = (2x-1)^2 + 5 = t^2 + 5$$, and the integral becomes $$\int \frac{t \cos\sqrt{t^2 + 5}}{\sqrt{t^2 + 5}} \cdot \frac{dt}{2}$$.

Now let $$u = \sqrt{t^2 + 5}$$, so $$u^2 = t^2 + 5$$ and $$2u\,du = 2t\,dt$$, giving $$t\,dt = u\,du$$. The integral becomes $$\frac{1}{2}\int \frac{u\cos u}{u}\,du = \frac{1}{2}\int \cos u\,du = \frac{1}{2}\sin u + c$$.

Substituting back: $$u = \sqrt{t^2 + 5} = \sqrt{(2x-1)^2 + 5}$$, so the answer is $$\frac{1}{2}\sin\sqrt{(2x-1)^2 + 5} + c$$.

The correct answer is (A): $$\frac{1}{2}\sin\sqrt{(2x-1)^2+5} + c$$.

We need to evaluate $$f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2}\, dx$$ for $$x \geq 0$$ with $$f(0) = 0$$. The key idea is to manipulate the integrand into a form $$-du/u^2$$, whose antiderivative is $$1/u$$.

Observe that the denominator is $$(x^2 + 1 + 2x^7)^2$$. The highest power of $$x$$ inside the bracket is $$x^7$$, so $$x^{14}$$ is the highest power in the denominator. We divide both numerator and denominator by $$x^{14}$$ (valid for $$x > 0$$). The numerator becomes $$\frac{5x^8 + 7x^6}{x^{14}} = 5x^{-6} + 7x^{-8}$$. The denominator becomes $$\left(\frac{x^2 + 1 + 2x^7}{x^7}\right)^2 = \left(x^{-5} + x^{-7} + 2\right)^2$$.

Now set $$u = x^{-5} + x^{-7} + 2$$. Differentiating gives $$\frac{du}{dx} = -5x^{-6} - 7x^{-8}$$, which is exactly $$-(5x^{-6} + 7x^{-8})$$, the negative of our transformed numerator. So the integral becomes $$\int \frac{-\,du}{u^2} = \frac{1}{u} + C$$.

Substituting back: $$f(x) = \frac{1}{x^{-5} + x^{-7} + 2} + C = \frac{x^7}{x^2 + 1 + 2x^7} + C$$ for $$x > 0$$. As $$x \to 0^+$$, the expression $$\frac{x^7}{x^2 + 1 + 2x^7} \to \frac{0}{0 + 1 + 0} = 0$$, so by continuity and the condition $$f(0) = 0$$, we get $$C = 0$$.

Finally, $$f(1) = \frac{1^7}{1^2 + 1 + 2 \cdot 1^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$$. Since $$f(1) = \frac{1}{K}$$, we conclude $$K = 4$$.

Therefore the answer is $$\boxed{4}$$.

Step 1: Simplify the integrand using algebraic manipulation

The given integral is:

$$I(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$$

We can rewrite the expression in the denominator by multiplying and dividing the second term by $$(x-11)^{\frac{15}{13}}$$:

$$(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}} = (x-11)^{\frac{11}{13}} \cdot (x-11)^{\frac{15}{13}} \cdot \left(\frac{x+15}{x-11}\right)^{\frac{15}{13}}$$

$$= (x-11)^{\frac{26}{13}} \cdot \left(\frac{x+15}{x-11}\right)^{\frac{15}{13}} = (x-11)^2 \cdot \left(\frac{x+15}{x-11}\right)^{\frac{15}{13}}$$

Substituting this back into the integral:

$$I(x)=\int \frac{1}{(x-11)^2} \cdot \left(\frac{x-11}{x+15}\right)^{\frac{15}{13}} d x$$

Step 2: Choose a suitable substitution

Let $$t = \frac{x+15}{x-11}$$.

Now, differentiate $$t$$with respect to$$x$$ using the quotient rule:

$$\frac{dt}{dx} = \frac{(x-11)(1) - (x+15)(1)}{(x-11)^2} = \frac{-26}{(x-11)^2}$$

$$\implies \frac{dx}{(x-11)^2} = -\frac{1}{26} dt$$

Step 3: Integrate with respect to $$t$$

Substitute the terms into the integral expression:

$$I(x) = \int t^{-\frac{15}{13}} \left( -\frac{1}{26} \right) dt = -\frac{1}{26} \int t^{-\frac{15}{13}} dt$$

Using the power rule of integration $$\int t^n dt = \frac{t^{n+1}}{n+1}$$:

$$I(x) = -\frac{1}{26} \left[ \frac{t^{-\frac{15}{13} + 1}}{-\frac{15}{13} + 1} \right] + C$$

$$I(x) = -\frac{1}{26} \left[ \frac{t^{-\frac{2}{13}}}{-\frac{2}{13}} \right] + C = -\frac{1}{26} \cdot \left(-\frac{13}{2}\right) t^{-\frac{2}{13}} + C$$

$$I(x) = \frac{1}{4} t^{-\frac{2}{13}} + C$$

Substitute back $$t = \frac{x+15}{x-11}$$:

$$I(x) = \frac{1}{4} \left(\frac{x+15}{x-11}\right)^{-\frac{2}{13}} + C = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}} + C$$

Step 4: Compute $$I(37) - I(24)$$

Now evaluate the definite difference:

For $$x = 37$$:

$$\frac{37-11}{37+15} = \frac{26}{52} = \frac{1}{2}$$

$$I(37) = \frac{1}{4} \left(\frac{1}{2}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{1}{4}\right)^{\frac{1}{13}} + C$$

For $$x = 24$$:

$$\frac{24-11}{24+15} = \frac{13}{39} = \frac{1}{3}$$

$$I(24) = \frac{1}{4} \left(\frac{1}{3}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{1}{9}\right)^{\frac{1}{13}} + C$$

Subtracting the two values:

$$I(37) - I(24) = \frac{1}{4} \left[ \left(\frac{1}{4}\right)^{\frac{1}{13}} - \left(\frac{1}{9}\right)^{\frac{1}{13}} \right] = \frac{1}{4} \left( \frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}} \right)$$

Step 5: Compare coefficients to find the final value

Comparing this result to the given equation $$\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right)$$:

$$b = 4$$and$$c = 9$$

Thus, the value of the required expression is:

$$3(b+c) = 3(4+9) = 3(13) = 39$$

1. Simplify the Integrand
To evaluate the integral, divide both the numerator and the denominator by $$x^2$$:
$$\int \frac{x^2+1}{x^4+1} \, dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx$$

2. Rewrite the Denominator
We can express the denominator $$x^2 + \frac{1}{x^2}$$ in terms of the algebraic identity $$(x - \frac{1}{x})^2$$:
$$\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \implies x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$$

Substitute this back into our integral expression:
$$\int \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 2} \, dx$$

3. Use Substitution Method
Let us use the substitution method by setting:
$$u = x - \frac{1}{x}$$

Differentiating both sides with respect to $$x$$ gives:
$$du = \left(1 + \frac{1}{x^2}\right) dx$$

Substitute $$u$$ and $$du$$ into the rewritten integral:
$$\int \frac{1}{u^2 + 2} \, du = \int \frac{1}{u^2 + (\sqrt{2})^2} \, du$$

4. Evaluate the Integral
Using the standard integration formula $$\int \frac{1}{y^2 + a^2} \, dy = \frac{1}{a} \tan^{-1}\left(\frac{y}{a}\right) + C$$, we get:
$$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$$

5. Substitute Back the Original Variable
Replace $$u$$ with its original definition $$x - \frac{1}{x}$$:
$$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{2}x}\right) + C$$

Let the given integral be denoted by $$I = \int \frac{x^2 - 1}{x \sqrt{x^4 + 3x^2 + 1}} \, dx$$.

To solve this integration, we can manipulate the algebraic terms by dividing both the numerator and the denominator by $$x^2$$ to reveal a hidden reciprocating functional group:
$$I = \int \frac{\frac{x^2 - 1}{x^2}}{\frac{x}{x^2} \sqrt{x^4 + 3x^2 + 1}} \, dx$$
$$I = \int \frac{1 - \frac{1}{x^2}}{\frac{1}{x} \sqrt{x^4 + 3x^2 + 1}} \, dx$$

Now, pull the factor of $$\frac{1}{x}$$ from the denominator inside the square root by rewriting it as $$\sqrt{\frac{1}{x^2}}$$ under the assumption that $$x > 0$$ for the valid function domain:
$$I = \int \frac{1 - \frac{1}{x^2}}{\sqrt{\frac{1}{x^2} \left(x^4 + 3x^2 + 1\right)}} \, dx$$
$$I = \int \frac{1 - \frac{1}{x^2}}{\sqrt{x^2 + 3 + \frac{1}{x^2}}} \, dx$$

Let us apply the method of substitution by using a helper variable $$u = x + \frac{1}{x}$$. Differentiating both sides with respect to $$x$$ gives:
$$du = \left(1 - \frac{1}{x^2}\right) dx$$

Notice that this differential expression matches our numerator perfectly. Next, express the polynomial expression inside the square root in terms of $$u$$ by squaring our substitution definition:
$$u^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$
$$x^2 + \frac{1}{x^2} = u^2 - 2$$

Substitute this equivalence back into the radical expression:
$$x^2 + 3 + \frac{1}{x^2} = \left(x^2 + \frac{1}{x^2}\right) + 3 = (u^2 - 2) + 3 = u^2 + 1$$

Now, substitute $$u$$ and $$du$$ into the transformed integral:
$$I = \int \frac{1}{\sqrt{u^2 + 1}} \, du$$

This matches the standard primary logarithmic integration formula $$\int \frac{1}{\sqrt{t^2 + a^2}} \, dt = \ln \left| t + \sqrt{t^2 + a^2} \right| + C$$. Applying it here with $$a = 1$$:
$$I = \ln \left| u + \sqrt{u^2 + 1} \right| + C$$

Finally, substitute back the original value for $$u = x + \frac{1}{x} = \frac{x^2 + 1}{x}$$ to write the final solution expression:
$$I = \ln \left| \frac{x^2 + 1}{x} + \sqrt{x^2 + 3 + \frac{1}{x^2}} \right| + C$$

Let the given integral be denoted as $$I = \int \frac{\sin 2x}{\cos^4 x + \sin^4 x} \, dx$$.

We can solve this problem efficiently by rewriting the denominator in terms of double-angle trigonometric functions to match the structure of the numerator.

Step 1: Simplify the denominator using algebraic identities
Recall the identity for the sum of squares, $$A^2 + B^2 = (A + B)^2 - 2AB$$. Let $$A = \cos^2 x$$ and $$B = \sin^2 x$$:
$$\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x$$

Using the fundamental identity $$\cos^2 x + \sin^2 x = 1$$, this simplifies to:
$$\cos^4 x + \sin^4 x = 1^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$$

Multiply and divide the second term by 2 to construct the double-angle identity $$2\sin x \cos x = \sin 2x$$:
$$\cos^4 x + \sin^4 x = 1 - \frac{4\sin^2 x \cos^2 x}{2} = 1 - \frac{(2\sin x \cos x)^2}{2} = 1 - \frac{\sin^2 2x}{2}$$

Now, express $$\sin^2 2x$$ in terms of $$\cos 2x$$ using the identity $$\sin^2 2x = 1 - \cos^2 2x$$:
$$\cos^4 x + \sin^4 x = 1 - \frac{1 - \cos^2 2x}{2} = 1 - \frac{1}{2} + \frac{\cos^2 2x}{2} = \frac{1 + \cos^2 2x}{2}$$

Step 2: Substitute this back into the integral
$$I = \int \frac{\sin 2x}{\frac{1 + \cos^2 2x}{2}} \, dx = \int \frac{2\sin 2x}{1 + \cos^2 2x} \, dx$$

Step 3: Apply substitution to evaluate the integral
Let us substitute a new variable $$u = \cos 2x$$. Differentiating both sides with respect to $$x$$ gives:
$$du = -2\sin 2x \, dx \implies 2\sin 2x \, dx = -du$$

Substitute $$u$$ and $$-du$$ into the transformed integral:
$$I = \int \frac{-1}{1 + u^2} \, du$$

This matches the standard primary integration formula $$\int \frac{1}{1 + t^2} \, dt = \tan^{-1} t$$:
$$I = -\tan^{-1}(u) + C$$

Step 4: Substitute back the original variable value
Replace $$u$$ with $$\cos 2x$$ to get the final solution:
$$I = -\tan^{-1}(\cos 2x) + C$$