JEE Mathematics Definite Integration MCQs Set 39

Answer: (c) \(\frac\pi4\)

Explanation:
1. \(\cos^2x=\frac{1+\cos2x}{2}\).
2. \(\int_0^{\pi/2}\frac{1+\cos2x}{2}dx=\frac12\left[x+\frac{\sin2x}{2}\right]_0^{\pi/2}=\frac\pi4\).

Teacher's Note:
1. Use the half-angle form of \(\cos^2x\) for any limits.

Answer: (b) \(1+\frac4\pi-\frac{2\sqrt2}{\pi}\)

Explanation:
1. \(\int_0^1xe^xdx=[(x-1)e^x]_0^1=0-(-1)=1\).
2. \(\int_0^1\sin\frac{\pi x}{4}dx=\frac4\pi\left[-\cos\frac{\pi x}{4}\right]_0^1=\frac4\pi\left(1-\frac{1}{\sqrt2}\right)=\frac4\pi-\frac{2\sqrt2}{\pi}\).
3. The sum is \(1+\frac4\pi-\frac{2\sqrt2}{\pi}\).

Teacher's Note:
1. Evaluate each part separately. The antiderivative of \(xe^x\) is \((x-1)e^x\).

Answer: (c) \(\frac\pi8\)

Explanation:
1. \(x^2+2x+5=(x+1)^2+4\), so the integral is \(\frac12\tan^{-1}\frac{x+1}{2}\).
2. \(\frac12\left[\tan^{-1}1-\tan^{-1}0\right]=\frac12\cdot\frac\pi4=\frac\pi8\).

Teacher's Note:
1. Complete the square before using the arctangent formula.

Answer: (a) \(\frac{e^2}{2}\left(\frac{e^2}{2}-1\right)\)

Explanation:
1. Let \(f(x)=\frac{e^{2x}}{2x}\). Then \(f'(x)=\frac{e^{2x}}{x}-\frac{e^{2x}}{2x^2}=\left(\frac1x-\frac1{2x^2}\right)e^{2x}\).
2. The integral is \(f(2)-f(1)=\frac{e^4}{4}-\frac{e^2}{2}=\frac{e^2}{2}\left(\frac{e^2}{2}-1\right)\).

Teacher's Note:
1. Look for an antiderivative of the form \(e^{2x}g(x)\). Here \(g=rac1{2x}\).

Answer: (c) \(e^{\pi/2}\)

Explanation:
1. \(\frac{1-\sin x}{1-\cos x}=\frac12\csc^2\frac x2-\cot\frac x2\). With \(f=-\cot\frac x2\), \(f'=\frac12\csc^2\frac x2\).
2. By \(\int e^x(f+f')=e^xf\), the integral is \(\left[-e^x\cot\frac x2\right]_{\pi/2}^{\pi}\).
3. \(=0-(-e^{\pi/2}\cdot1)=e^{\pi/2}\).

Teacher's Note:
1. Use \(1-\cos x=2\sin^2rac x2\) and \(1-\sin x\) in half angles to expose the \(e^x(f+f')\) form.

Answer: (b) \(\frac{\pi^2-4\pi}{16}+\log\sqrt2\)

Explanation:
1. By parts, \(\int x(\tan^{-1}x)^2dx=\frac{x^2+1}{2}(\tan^{-1}x)^2-\int\tan^{-1}x\,dx\).
2. Evaluating between 0 and 1: \(\frac{\pi^2}{16}-\left[x\tan^{-1}x-\frac12\log(1+x^2)\right]_0^1=\frac{\pi^2}{16}-\frac\pi4+\frac12\log2\).
3. This equals \(\frac{\pi^2-4\pi}{16}+\log\sqrt2\) (checked numerically: 0.1780).

Teacher's Note:
1. Choose \(rac{x^2+1}{2}\) as the antiderivative of x to simplify the remaining integral.

Answer: (b) \(a+1-\frac e2\)

Explanation:
1. By parts: \(\int_0^1\frac{e^t}{(1+t)^2}dt=\left[-\frac{e^t}{1+t}\right]_0^1+\int_0^1\frac{e^t}{1+t}dt\).
2. \(=-\frac e2+1+a\).
3. So the value is \(a+1-\frac e2\).

Teacher's Note:
1. Integrate \(rac1{(1+t)^2}\) as the second function. The leftover integral is the given one.

Answer: (b) 3

Explanation:
1. By parts, \(\int_0^\pi f''\sin x\,dx=[f'\sin x]_0^\pi-\int_0^\pi f'\cos x\,dx=-\int_0^\pi f'\cos x\,dx\).
2. \(\int_0^\pi f'\cos x\,dx=[f\cos x]_0^\pi+\int_0^\pi f\sin x\,dx=-f(\pi)-f(0)+\int_0^\pi f\sin x\,dx\).
3. Hence \(\int(f+f'')\sin x=f(\pi)+f(0)=5\), so \(f(0)=3\).

Teacher's Note:
1. Apply integration by parts twice; the \(\int f\sin x\) terms cancel.

Answer: (b) \(x\sin x\)

Explanation:
1. By the Fundamental Theorem of Calculus, \(\frac{d}{dx}\int_0^xg(t)dt=g(x)\).
2. With \(g(t)=t\sin t\), \(f'(x)=x\sin x\).

Teacher's Note:
1. Do not integrate; just replace t by x in the integrand.

Answer: (c) \(\frac\pi4\)

Explanation:
1. Let \(I\) be the integral. Replace \(x\) by \(\frac\pi2-x\) to get \(I=\int_0^{\pi/2}\frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}dx\).
2. Adding, \(2I=\int_0^{\pi/2}1\,dx=\frac\pi2\), so \(I=\frac\pi4\).

Teacher's Note:
1. The property \(\int_0^af(x)dx=\int_0^af(a-x)dx\) makes the sum of the two forms equal 1.

Answer: (a) \(\pi\)

Explanation:
1. Using \(x\to\pi-x\): \(I=\pi\int_0^\pi\frac{dx}{1+\sin x}-I\), so \(I=\frac\pi2\int_0^\pi\frac{dx}{1+\sin x}\).
2. By symmetry about \(\frac\pi2\), \(\int_0^\pi\frac{dx}{1+\sin x}=2\int_0^{\pi/2}\frac{dx}{1+\sin x}=2\left[\tan x-\sec x\right]_0^{\pi/2}=2(0-(-1))=2\).
3. So \(I=\frac\pi2\cdot2=\pi\).

Teacher's Note:
1. For \(\int x f(\sin x)\) over \([0,\pi]\) use \(x o\pi-x\) to reach \(rac\pi2\int f(\sin x)\).

Answer: (d) \(0\)

Explanation:
1. \(\cos^5(2\pi-x)=\cos^5x\), so the integral is \(2\int_0^\pi\cos^5x\,dx\).
2. \(\cos^5(\pi-x)=-\cos^5x\), so \(\int_0^\pi\cos^5x\,dx=0\).
3. Hence the value is 0.

Teacher's Note:
1. Odd powers of cosine over a full period integrate to zero.

Answer: (c) \(\frac\pi4\)

Explanation:
1. Replacing \(x\) by \(\frac\pi2-x\) gives \(I=\int_0^{\pi/2}\frac{\sin^5x}{\sin^5x+\cos^5x}dx\).
2. Adding, \(2I=\frac\pi2\), so \(I=\frac\pi4\).

Teacher's Note:
1. Same property as before: the two forms add up to 1.

Answer: (a) \(0\)

Explanation:
1. Replace \(x\) by \(\frac\pi2-x\): the numerator changes sign and the denominator is unchanged.
2. So \(I=-I\), hence \(I=0\).

Teacher's Note:
1. If \(f(a-x)=-f(x)\), then \(\int_0^af(x)dx=0\).

Answer: (d) \(\frac a2\)

Explanation:
1. Replacing \(x\) by \(a-x\) gives \(I=\int_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt x}dx\).
2. Adding, \(2I=\int_0^a1\,dx=a\), so \(I=\frac a2\).

Teacher's Note:
1. The King property \(\int_0^af(x)dx=\int_0^af(a-x)dx\) solves this at once.

Answer: (d) \(5\)

Explanation:
1. Split at \(x=1\): \(\int_0^1(1-x)dx=\frac12\) and \(\int_1^4(x-1)dx=\frac92\).
2. The sum is 5.

Teacher's Note:
1. Split the interval at the point where the expression inside the modulus is zero.

Answer: (c) 0

Explanation:
1. Let \(I=\int_0^{\pi/2}\log(4+3\sin x)dx-\int_0^{\pi/2}\log(4+3\cos x)dx\).
2. By the substitution \(x\to\frac\pi2-x\), the two integrals are equal.
3. So the value is 0.

Teacher's Note:
1. The two logarithm integrals are the same function reflected about \(rac\pi4\).

Answer: (c) 0

Explanation:
1. \(f(x)=x^{17}\cos^4x\) satisfies \(f(-x)=-f(x)\), so it is an odd function.
2. The integral of an odd function over \([-a,a]\) is 0.

Teacher's Note:
1. Check for odd symmetry before integrating over a symmetric interval.

Answer: (d) \(a\pi\)

Explanation:
1. Multiply and divide by \(\sqrt{a-x}\): \(\int_{-a}^a\frac{a-x}{\sqrt{a^2-x^2}}dx\).
2. The term with \(x\) is odd and gives 0. So the value is \(a\int_{-a}^a\frac{dx}{\sqrt{a^2-x^2}}=a[\sin^{-1}\frac xa]_{-a}^a=a\pi\).

Teacher's Note:
1. Rationalise and then split into odd and even parts.

Answer: (a) \(0\)

Explanation:
1. \(x^3e^{x^4}\) is an odd function.
2. So its integral over \([-1,1]\) is 0.

Teacher's Note:
1. An odd integrand over a symmetric interval always gives zero.

Answer: (b) \(\frac13(4\sqrt2-1)\)

Explanation:
1. \(\int_0^1x^2dx=\frac13\).
2. \(\int_1^2\sqrt x\,dx=\frac23\left(2\sqrt2-1\right)\).
3. The total is \(\frac13+\frac{4\sqrt2-2}{3}=\frac{4\sqrt2-1}{3}\).

Teacher's Note:
1. Integrate each branch of a piecewise function over its own interval.

Answer: (d) \(\frac{28}{3}\)

Explanation:
1. \(|x^2-1|=x^2-1\) on \([-2,-1]\) and \([1,3]\), and \(1-x^2\) on \([-1,1]\).
2. \(\int_{-2}^{-1}(x^2-1)dx=\frac43\), \(\int_{-1}^1(1-x^2)dx=\frac43\), \(\int_1^3(x^2-1)dx=\frac{20}{3}\).
3. The sum is \(\frac{28}{3}\).

Teacher's Note:
1. Find where \(x^2-1\) changes sign, and split there.

Answer: (a) \(\frac{1}{\sqrt2}\log(\sqrt2+1)\)

Explanation:
1. By \(x\to\frac\pi2-x\), \(I=\int\frac{\cos^2x}{\sin x+\cos x}dx\). Adding, \(2I=\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}\).
2. \(\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right)\), so \(\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}=\frac{1}{\sqrt2}\left[\log\tan\frac{x+\pi/4}{2}\right]_0^{\pi/2}=\frac{2}{\sqrt2}\log(\sqrt2+1)\).
3. Hence \(I=\frac1{\sqrt2}\log(\sqrt2+1)\).

Teacher's Note:
1. Combine with the reflected integral first; then use the sine of a sum.

Answer: (d) \(\frac{19}{2}\)

Explanation:
1. \(\int_{-1}^2|x+1|dx=\frac92\); \(\int_{-1}^2|x|dx=\frac12+2=\frac52\); \(\int_{-1}^2|x-1|dx=2+\frac12=\frac52\).
2. The sum is \(\frac92+\frac52+\frac52=\frac{19}{2}\).

Teacher's Note:
1. Integrate each modulus term over \([-1,2]\), splitting at its own zero.

Answer: (c) 9

Explanation:
1. \(\frac{dx}{dy}=\frac{1}{\sqrt{1+9y^2}}\), so \(\frac{dy}{dx}=\sqrt{1+9y^2}\).
2. \(\frac{d^2y}{dx^2}=\frac{9y}{\sqrt{1+9y^2}}\cdot\frac{dy}{dx}=9y\).
3. So \(a=9\).

Teacher's Note:
1. Differentiate the integral limit using the Fundamental Theorem. Then use the chain rule once more.

Answer: (b) \(2\log2\)

Explanation:
1. The term \(\frac{x^3}{x^2+2|x|+1}\) is odd and integrates to 0.
2. The rest is even: \(2\int_0^1\frac{x+1}{(x+1)^2}dx=2\int_0^1\frac{dx}{x+1}=2\log2\).

Teacher's Note:
1. Split into odd and even parts. Note \(x^2+2|x|+1=(|x|+1)^2\).