JEE Mathematics Definite Integration MCQs Set 18

Answer: (d) \( 2\pi \)

Explanation:
1. Expand: \( (\cos ax-\sin bx)^2=\cos^2ax+\sin^2bx-2\cos ax\sin bx \).
2. The product \( \cos ax\sin bx \) is odd, so its integral over \( [-\pi,\pi] \) is 0.
3. \( \int_{-\pi}^{\pi}\cos^2ax\,dx=\pi \) and \( \int_{-\pi}^{\pi}\sin^2bx\,dx=\pi \) for non-zero integers a, b. Total \( 2\pi \).

Teacher's Note:
1. Odd functions integrate to zero over a symmetric interval.
2. Over a full period, \( \cos^2 \) and \( \sin^2 \) average to one half.

Answer: (b) \( \pi\log2 \)

Explanation:
1. Write \( \int x^2\csc^2x\,dx \) and integrate by parts: \( \left[-x^2\cot x\right]_0^{\pi/2}+\int_0^{\pi/2}2x\cot x\,dx \).
2. The boundary term is 0 at both ends.
3. \( \int_0^{\pi/2}x\cot x\,dx=\frac\pi2\ln2 \), so the integral is \( 2\cdot\frac\pi2\ln2=\pi\ln2 \).

Teacher's Note:
1. Use \( \csc^2x \) as the part that integrates directly.
2. The remaining integral of \( x\cot x \) is a standard result.

Answer: (c) \( F(t)=e^t-(1+t) \)

Explanation:
1. \( F(t)=\int_0^te^{t-y}y\,dy=e^t\int_0^tye^{-y}dy \).
2. \( \int_0^tye^{-y}dy=\left[-ye^{-y}-e^{-y}\right]_0^t=1-e^{-t}(1+t) \).
3. So \( F(t)=e^t-(1+t) \).

Teacher's Note:
1. Take the factor \( e^t \) outside because it does not depend on y.
2. Integrate \( ye^{-y} \) by parts.

Answer: (d) \( -\frac25\log2+\frac12 \)

Explanation:
1. Replacing x by \( \frac1x \): \( 2f\left(\frac1x\right)+3f(x)=x-2 \).
2. Eliminate \( f(1/x) \): \( 5f(x)=3x-2-\frac2x \), so \( f(x)=\frac15\left(3x-2-\frac2x\right) \).
3. \( \int_1^2f=\frac15\left[\frac{3x^2}{2}-2x-2\ln x\right]_1^2=\frac15\left(\frac52-2\ln2\right)=\frac12-\frac25\ln2 \).

Teacher's Note:
1. A functional equation involving x and 1/x is solved by writing it twice.
2. Then integrate the explicit function.

Answer: (a) 0

Explanation:
1. On \( [-1,1] \), \( x\sin\pi x\ge0 \) and it is even, with maximum about 0.57 near \( x=0.6 \).
2. So \( 0\le x\sin\pi x<1 \) and the greatest integer part is 0 throughout (the value would reach 1 only if the function did).
3. The integral is 0.

Teacher's Note:
1. Estimate the maximum value of \( x\sin\pi x \) on [0, 1].
2. If the function never reaches 1, its integer part is 0.

Answer: (b) \( \frac\pi2 \)

Explanation:
1. Put \( u=x+\pi \), so u runs from \( -\frac\pi2 \) to \( \frac\pi2 \) and \( \cos^2(x+3\pi)=\cos^2(u+2\pi)=\cos^2u \).
2. \( \int_{-\pi/2}^{\pi/2}u^3du=0 \) (odd function).
3. \( \int_{-\pi/2}^{\pi/2}\cos^2u\,du=\frac\pi2 \). The value is \( \frac\pi2 \).

Teacher's Note:
1. Shift the variable to centre the interval at 0.
2. The cube is odd and drops out.

Answer: (d) \( \frac3\pi+\frac1{\pi^2} \)

Explanation:
1. \( x\sin\pi x\ge0 \) on \( [-1,1] \) and \( \le0 \) on \( \left[1,\frac32\right] \). So the integral is \( \int_{-1}^1x\sin\pi x\,dx-\int_1^{3/2}x\sin\pi x\,dx \).
2. An antiderivative of \( x\sin\pi x \) is \( F(x)=-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^2} \). Then \( F(1)=\frac1\pi \), \( F(-1)=-\frac1\pi \), \( F\left(\frac32\right)=-\frac1{\pi^2} \).
3. \( \int_{-1}^1=\frac2\pi \) and \( \int_1^{3/2}=-\frac1{\pi^2}-\frac1\pi \). The total is \( \frac2\pi+\frac1\pi+\frac1{\pi^2}=\frac3\pi+\frac1{\pi^2} \).

Teacher's Note:
1. Split the interval where the sign of \( x\sin\pi x \) changes.
2. Integrate by parts for the antiderivative.

Answer: (b) \( 4\log\frac43 \)

Explanation:
1. The expression under the root is \( \left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^2=\left(\frac{4x}{x^2-1}\right)^2 \), so the integrand is \( \frac{4|x|}{1-x^2} \).
2. The integrand is even: \( 2\int_0^{1/2}\frac{4x}{1-x^2}dx=2\left[-2\ln(1-x^2)\right]_0^{1/2} \).
3. \( =2\cdot2\ln\frac43=4\ln\frac43 \).

Teacher's Note:
1. Recognise the perfect square \( a^2+b^2-2 \) with \( ab=1 \).
2. Use symmetry because the integrand is even.

Answer: (b) \( -\frac\pi3 \)

Explanation:
1. \( \int_x^{2x}\cos2t\,dt=\frac{\sin4x-\sin2x}{2} \), so the equation becomes \( 2\sin x+\sin4x-\sin2x=0 \).
2. Check \( x=-\frac\pi3 \): \( 2\left(-\frac{\sqrt3}{2}\right)+\sin\left(-\frac{4\pi}{3}\right)-\sin\left(-\frac{2\pi}{3}\right)=-\sqrt3+\frac{\sqrt3}{2}+\frac{\sqrt3}{2}=0 \).
3. The other options give non-zero values (for example \( x=-\frac\pi2 \) gives -2 on the left side of the original equation). So \( x=-\frac\pi3 \).

Teacher's Note:
1. Evaluate the integral first, then test the given values.
2. Use \( \sin(-\theta)=-\sin\theta \).

Answer: (d) 30

Explanation:
1. For non-integer x, \( [x-4]=[x]-4 \) and \( [10-x]=10+[-x]=10-[x]-1=9-[x] \).
2. Their sum is 5 (except at integers, which do not affect the integral).
3. \( \int_4^{10}5\,dx=5\times6=30 \).

Teacher's Note:
1. Use \( [-x]=-[x]-1 \) for non-integer x.
2. Isolated points do not change an integral.

Answer: (d) \( -\frac38 \)

Explanation:
1. \( f'(x)=|x|\ge0 \), so f is increasing on the interval.
2. The greatest value is \( f\left(\frac12\right)=\int_1^{1/2}t\,dt=\frac12\left(\frac14-1\right)=-\frac38 \).

Teacher's Note:
1. The derivative of the integral is the integrand.
2. An increasing function is greatest at the right end.

Answer: (b) \( \frac{4-\pi}{4+\pi} \)

Explanation:
1. Note that \( \dfrac{d}{dx}\left(\dfrac{\sin x-x\cos x}{x\sin x+\cos x}\right)=\dfrac{x^2}{(x\sin x+\cos x)^2} \).
2. At \( x=\frac\pi4 \): \( \dfrac{\frac{\sqrt2}{2}\left(1-\frac\pi4\right)}{\frac{\sqrt2}{2}\left(1+\frac\pi4\right)}=\dfrac{4-\pi}{4+\pi} \). At x = 0 the value is 0.
3. The integral equals \( \frac{4-\pi}{4+\pi} \).

Teacher's Note:
1. Look for a quotient whose derivative produces \( x^2 \) in the numerator.
2. Substitute the limits carefully.

Answer: (c) \( -\frac\pi2 \)

Explanation:
1. On \( \left[\frac\pi2,\frac{5\pi}6\right] \), \( 2\sin x\in[1,2] \): [.] = 1 (length \( \frac\pi3 \)). On \( \left(\frac{5\pi}6,\pi\right] \): \( 2\sin x\in[0,1) \): 0.
2. On \( \left(\pi,\frac{7\pi}6\right] \): \( 2\sin x\in[-1,0) \): -1 (length \( \frac\pi6 \)). On \( \left(\frac{7\pi}6,\frac{3\pi}2\right] \): \( 2\sin x\in[-2,-1) \): -2 (length \( \frac\pi3 \)).
3. Total \( =\frac\pi3-\frac\pi6-\frac{2\pi}3=-\frac\pi2 \).

Teacher's Note:
1. Find where \( 2\sin x \) crosses the integers 1, 0, -1.
2. Add the contributions of each strip.

Answer: (d) \( 4^n \)

Explanation:
1. The integrand is a constant: the sum of even-indexed binomial coefficients of \( (1+x)^{2n} \) is \( 2^{2n-1} \).
2. \( \int_0^22^{2n-1}dx=2\cdot2^{2n-1}=2^{2n}=4^n \).

Teacher's Note:
1. The sum of even-numbered binomial coefficients is half of \( 2^{2n} \).
2. The integral of a constant is constant times length.

Answer: (d) 1

Explanation:
1. \( \cos3x+1=4\cos^3x-3\cos x+1=(2\cos x-1)(2\cos^2x+\cos x-1) \).
2. So the integrand is \( 2\cos^2x+\cos x-1=\cos2x+\cos x \).
3. \( \int_0^{\pi/2}(\cos2x+\cos x)dx=\left[\frac{\sin2x}{2}+\sin x\right]_0^{\pi/2}=0+1=1 \).

Teacher's Note:
1. Factorise the numerator using the triple-angle formula.
2. After cancelling, the integral is elementary.

Answer: (c) 17

Explanation:
1. \( [x-1]=[x]-1 \). \( \int_0^{7.5}[x]\,dx=(0+1+2+\dots+6)+0.5\times7=21+3.5=24.5 \).
2. Subtract \( \int_0^{7.5}1\,dx=7.5 \): \( 24.5-7.5=17 \).

Teacher's Note:
1. Split the interval into unit steps where the integer part is constant.
2. The last half step has [x] = 7.

Answer: (b) \( 1-\sqrt3 \)

Explanation:
1. \( \int_a^bf''(x)dx=f'(b)-f'(a) \).
2. \( f'(a)=\tan\frac\pi3=\sqrt3 \) and \( f'(b)=\tan\frac\pi4=1 \).
3. The integral is \( 1-\sqrt3 \).

Teacher's Note:
1. The slope of a tangent equals the tangent of its inclination.
2. Use the fundamental theorem of calculus.

Answer: (c) \( y\sqrt2=x-1 \)

Explanation:
1. At x = 1 both limits equal 1, so \( y=0 \): the point is (1, 0).
2. \( \dfrac{dy}{dx}=\dfrac{3x^2}{\sqrt{1+x^6}}-\dfrac{2x}{\sqrt{1+x^4}}=\dfrac3{\sqrt2}-\dfrac2{\sqrt2}=\dfrac1{\sqrt2} \) at x = 1.
3. Tangent: \( y=\frac{x-1}{\sqrt2} \), i.e. \( y\sqrt2=x-1 \).

Teacher's Note:
1. Differentiate under the integral sign with variable limits (Leibniz rule).
2. The point of contact comes from equal limits.

Answer: (d) \( e-\frac{e^2}{2}-\frac32 \)

Explanation:
1. \( f'=f \) with f(0) = 1 gives \( f(x)=e^x \), so \( g(x)=x^2-e^x \).
2. \( \int_0^1e^x(x^2-e^x)dx=\int_0^1x^2e^xdx-\int_0^1e^{2x}dx \).
3. \( \int_0^1x^2e^xdx=\left[e^x(x^2-2x+2)\right]_0^1=e-2 \) and \( \int_0^1e^{2x}dx=\frac{e^2-1}2 \). The value is \( e-2-\frac{e^2}2+\frac12=e-\frac{e^2}{2}-\frac32 \).

Teacher's Note:
1. Solve the differential equation first.
2. Integrate \( x^2e^x \) by parts twice.

Answer: (a) \( \mu-b \)

Explanation:
1. The slope function is \( \phi(x)=f'(x) \).
2. \( \int_a^\lambda f'(x)dx=f(\lambda)-f(a)=\mu-b \).

Teacher's Note:
1. The integral of a derivative is the change in the function.
2. The y-coordinates of the two points are b and \( \mu \).

Answer: (d) 3

Explanation:
1. \( f'(x)=\dfrac1{f(x)^2} \), so \( f^2df=dx \) and \( \frac{f^3}{3}=x+C \). Since \( f(0)=0 \), \( f(x)^3=3x \).
2. Check: \( f(2)=\sqrt[3]6 \), which matches the given integral.
3. \( f(9)=\sqrt[3]{27}=3 \).

Teacher's Note:
1. Differentiate to get a separable differential equation.
2. The initial value is f(0) = 0.

Answer: (c) \( \frac12+\frac1{\pi+2}-A \)

Explanation:
1. By parts: \( A=\left[-\frac{\cos x}{x+2}\right]_0^\pi-\int_0^\pi\frac{\sin x}{x+2}dx=\frac1{\pi+2}+\frac12-\int_0^\pi\frac{\sin x}{x+2}dx \).
2. In \( \int_0^{\pi/2}\frac{\sin2x}{x+1}dx \) put \( y=2x \): it becomes \( \int_0^\pi\frac{\sin y}{y+2}dy \).
3. Hence the required integral is \( \frac12+\frac1{\pi+2}-A \).

Teacher's Note:
1. Integrate A by parts to expose \( \int\frac{\sin x}{x+2}dx \).
2. A change of variable y = 2x turns the second integral into the same form.

Answer: (c) \( \sqrt2 \)

Explanation:
1. Replacing x by \( \frac\pi2-x \) leaves \( \sin2x \) unchanged, so \( I_1=\int_0^{\pi/2}f(\sin2x)\cos x\,dx \). Adding: \( 2I_1=\int_0^{\pi/2}f(\sin2x)(\sin x+\cos x)dx \).
2. The integrand is symmetric about \( x=\frac\pi4 \), so \( I_1=\int_0^{\pi/4}f(\sin2x)(\sin x+\cos x)\,dx \).
3. Put \( x=\frac\pi4-u \): \( \sin2x=\cos2u \) and \( \sin x+\cos x=\sqrt2\cos u \), so \( I_1=\sqrt2\int_0^{\pi/4}f(\cos2u)\cos u\,du=\sqrt2\,I_2 \).

Teacher's Note:
1. Use the symmetry \( x\to\frac\pi2-x \) to bring in cos x.
2. The substitution \( x=\frac\pi4-u \) converts \( \sin2x \) into \( \cos2u \).