Indefinite Integrals — JEE Main practice

1. Why: Combine the square-root factors in the denominator into a single expression, since $\sqrt{1+x}(1-x)=\sqrt{1-x^2}\cdot(1-x)$... actually simplify directly using $\sqrt{1+x}\cdot(1-x)^{3/2}=\sqrt{1-x^2}\,(1-x)$.

$$f(x)=\int\frac{(2-x^2)e^x}{\sqrt{1-x^2}\,(1-x)}\,dx$$

2. Why: Split the numerator $2-x^2$ as $1+(1-x^2)$ so the integral separates into two recognizable pieces.

$$f(x)=\int e^x\left[\frac{1}{\sqrt{1-x^2}(1-x)}+\frac{1+x}{\sqrt{1-x^2}}\right]dx$$

3. Why: This matches the standard form $\int e^x[g(x)+g'(x)]dx=e^xg(x)+C$; guess $g(x)=\frac{1+x}{\sqrt{1-x^2}}$ and verify by differentiating.

$$g'(x)=\frac{\sqrt{1-x^2}+\frac{x^2+x}{\sqrt{1-x^2}}}{1-x^2}=\frac{1}{\sqrt{1-x^2}(1-x)}$$

This matches the first bracket term, confirming the guess.

4. Why: Apply the standard-form result directly.

$$f(x)=\frac{e^x(1+x)}{\sqrt{1-x^2}}+C$$

5. Why: Use the given condition $f(0)=0$ to solve for $C$.

$$0=\frac{e^0(1)}{\sqrt1}+C \implies C=-1$$

6. Why: Evaluate $f\left(\frac12\right)$ using the completed formula.

$$f\left(\frac12\right)=\frac{e^{1/2}\cdot\frac32}{\sqrt{3}/2}-1=\sqrt{3e}-1$$

Hence, the answer is Option C.

1. Why: Rewrite the denominator by factoring out $2$ from $(4x+6)$ and completing the square inside the square root, to reveal a form suitable for substitution.

$$I(x)=\int\frac{3\,dx}{2(2x+3)\sqrt{(2x+2)^2-1}}$$

2. Why: Substitute $2x+3=\frac{1}{t}$ so the linear factor outside the root simplifies to $\frac1t$, converting the integral into a function of $t$.

$$2x+3=\frac1t \implies dx=-\frac{1}{2t^2}dt,\qquad 2x+2=\frac{1-t}{t}$$

3. Why: Substitute these into $I(x)$ and simplify the expression under the square root.

$$I(x)=\int\frac{-3\,dt}{4\sqrt{1-2t}}$$

4. Why: This is now a direct power-rule integral in $t$; integrate it.

$$I(x)=\frac34\sqrt{1-2t}+C$$

5. Why: Substitute back $t=\frac{1}{2x+3}$ to express $I$ in terms of $x$.

$$I(x)=\frac34\sqrt{\frac{2x+1}{2x+3}}+C$$

6. Why: Use the given value $I(0)=\frac{\sqrt3}{4}+20$ to solve for $C$.

$$\frac34\sqrt{\frac13}+C=\frac{\sqrt3}{4}+20 \implies \frac{\sqrt3}{4}+C=\frac{\sqrt3}{4}+20 \implies C=20$$

7. Why: Evaluate $I\left(\frac12\right)$ using the completed formula.

$$I\left(\frac12\right)=\frac34\sqrt{\frac{2}{4}}+20=\frac{3\sqrt2}{8}+20$$

Comparing with $\frac{a\sqrt2}{b}+c$: $a=3,\ b=8,\ c=20$.

8. Why: Compute the requested sum.

$$a+b+c=3+8+20=31$$

Hence, the answer is Option D.

1. Why: Substitute $x=\log_e t$ so that $t=e^x$ and the integral is written purely in terms of $x$ and $e^x$, since the argument of $\sin$ and $\cos$ is $\log_e t$.

$$f(t)=\int\left(\frac{1-\sin x}{1-\cos x}\right)e^x\,dx$$

2. Why: Use the half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1-\cos x=2\sin^2\frac{x}{2}$ to simplify the trig fraction.

$$f(t)=\int\left[\frac{1}{2}\operatorname{cosec}^2\frac{x}{2}-\cot\frac{x}{2}\right]e^x\,dx$$

3. Why: Recognize this as the standard form $\int e^x[g(x)+g'(x)]\,dx=e^x g(x)+C$; check that $g(x)=-\cot\frac{x}{2}$ has derivative matching the remaining term.

$$g'(x)=\frac{1}{2}\operatorname{cosec}^2\frac{x}{2}$$

So

$$f(t)=e^x\left(-\cot\frac{x}{2}\right)+C$$

4. Why: Substitute $x=\log_e t$ back to express $f$ in terms of $t$.

$$f(t)=-t\cot\left(\frac{\log_e t}{2}\right)+C$$

5. Why: Use the given condition $f(e^{\pi/2})=-e^{\pi/2}$ to solve for $C$.

$$-e^{\pi/2}=-e^{\pi/2}\cot\left(\frac{\pi}{4}\right)+C=-e^{\pi/2}+C \implies C=0$$

6. Why: With $C=0$, evaluate $f(e^{\pi/4})$ and equate it to $\alpha e^{\pi/4}$ to solve for $\alpha$.

$$\alpha e^{\pi/4}=-e^{\pi/4}\cot\left(\frac{\pi}{8}\right) \implies \alpha=-\cot\left(\frac{\pi}{8}\right)$$

7. Why: Use the identity $\cot\theta=\frac{1+\cos2\theta}{\sin2\theta}$ with $\theta=\frac{\pi}{8}$ to evaluate $\cot\frac{\pi}{8}$.

$$\alpha=-\left(\frac{1+\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}\right)=-\left(\frac{1+\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}\right)=-(\sqrt2+1)$$

Hence, the answer is Option C.