JEE Mathematics Indefinite Integration MCQs Set 04

Answer: (b) \( x^{5} - \frac{5 x^{3}}{3} + 5 x+c \)

Explanation:
1. \( x^6+1=(x^2+1)(x^4-x^2+1) \), so the integrand is \( 5(x^4-x^2+1) \).
2. Integrate: \( x^5-\frac53x^3+5x+c \).

Teacher's Note:
1. Sum of cubes factorisation x cubed plus 1 with x replaced by x squared.
2. After cancelling, the integrand is a polynomial.

Answer: (d) \( - \frac{\cos{\left(2 x \right)}}{4}+c \)

Explanation:
1. \( \tan x+\cot x=\frac{1}{\sin x\cos x} \), so the integrand is \( \sin x\cos x=\frac{\sin2x}{2} \).
2. \( \int\frac{\sin2x}{2}dx=-\frac{\cos2x}{4}+c \).

Teacher's Note:
1. Write tan and cot in terms of sine and cosine.
2. Use the double angle formula for sin 2x.

Answer: (a) \( \log{\left(x \right)} - 2 \cos{\left(x \right)}+c \)

Explanation:
1. \( \int2\sin x\,dx=-2\cos x \) and \( \int\frac1xdx=\log x \).
2. The result is \( -2\cos x+\log x+C \).

Teacher's Note:
1. Integrate term by term.
2. Option (d) confuses integration with differentiation.

Answer: (b) \( \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}+c \)

Explanation:
1. \( \sin2x\cos3x=\frac12(\sin5x-\sin x) \).
2. \( \frac12\left(-\frac{\cos5x}{5}+\cos x\right)+c=\frac12\left(\cos x-\frac15\cos5x\right)+c \).

Teacher's Note:
1. Convert the product into a sum with 2 sin A cos B.
2. Remember the factor 1/2.

Answer: (d) \( a=-\dfrac{5\pi}{4},\ b=\text{any constant} \)

Explanation:
1. \( \int(\sin2x-\cos2x)dx=-\frac12(\cos2x+\sin2x)+k=-\frac{1}{\sqrt2}\sin\left(2x+\frac\pi4\right)+k \).
2. Since \( -\sin\theta=\sin(\theta+\pi) \), this is \( \frac{1}{\sqrt2}\sin\left(2x+\frac{5\pi}{4}\right)+k \), so \( a=-\frac{5\pi}{4} \) and b is an arbitrary constant.

Teacher's Note:
1. Combine sin and cos into a single sine using R sin(theta + phi).
2. The constant of integration appears as b.

Answer: (a) \( c=\dfrac{\pi}{4} \) and \( a=k \) (an arbitrary constant)

Explanation:
1. \( \int(\sin2x+\cos2x)dx=\frac12(\sin2x-\cos2x)+k=\frac{1}{\sqrt2}\sin\left(2x-\frac\pi4\right)+k \).
2. So \( c=\frac\pi4 \) and a is the arbitrary constant.

Teacher's Note:
1. Write sin 2x minus cos 2x as root 2 times sin of 2x minus pi/4.
2. The constant a is just the constant of integration.

Answer: (b) \( -\dfrac{\cos2x}{4}+\text{constant} \)

Explanation:
1. \( \sin5x\cos3x=\frac12(\sin8x+\sin2x) \), so the integral is \( -\frac{\cos8x}{16}-\frac{\cos2x}{4}+k \).
2. Comparing, \( A=-\frac{\cos2x}{4}+k \).

Teacher's Note:
1. Product-to-sum conversion gives two terms.
2. The second term is absorbed into A.

Answer: (a) \( \left(\dfrac12,\dfrac\pi4\right) \)

Explanation:
1. \( \sqrt{1+\sin x}=\sin\frac x2+\cos\frac x2 \), so \( \sqrt2\sqrt{1+\sin x}=2\sin\left(\frac x2+\frac\pi4\right) \).
2. \( \int2\sin\left(\frac x2+\frac\pi4\right)dx=-4\cos\left(\frac x2+\frac\pi4\right)+C \), so \( (a,b)=\left(\frac12,\frac\pi4\right) \).

Teacher's Note:
1. Express sin plus cos as root 2 times sine of a shifted angle.
2. Integrating sin of x/2 gives minus 2 cos of x/2.

Answer: (a) \( \sin{\left(x \right)} - \csc{\left(x \right)}+c \)

Explanation:
1. \( \frac{\sin x}{\tan x}=\cos x \) and \( \frac{\operatorname{cosec}x}{\tan x}=\frac{\cos x}{\sin^2x}=\operatorname{cosec}x\cot x \).
2. \( \int(\cos x+\operatorname{cosec}x\cot x)dx=\sin x-\operatorname{cosec}x+c \).

Teacher's Note:
1. Split the fraction and simplify each part with sin and cos.
2. The integral of cosec x cot x is minus cosec x.

Answer: (c) \( \sqrt{2} \log{\left(\tan{\left(\frac{x}{4} + \frac{\pi}{8} \right)} \right)}+c \)

Explanation:
1. \( 1+\sin x=\left(\sin\frac x2+\cos\frac x2\right)^2=2\sin^2\left(\frac x2+\frac\pi4\right) \).
2. The integrand is \( \frac{1}{\sqrt2}\operatorname{cosec}\left(\frac x2+\frac\pi4\right) \), and \( \int\operatorname{cosec}u\,du=\log\tan\frac u2 \).
3. Hence the integral is \( \sqrt2\log\tan\left(\frac x4+\frac\pi8\right)+c \).

Teacher's Note:
1. Write 1 + sin x as 2 sin squared of (x/2 + pi/4).
2. Dividing by the coefficient 1/2 gives the factor 2 over root 2.

Answer: (a) \( \tan{\left(x \right)} + \cot{\left(x \right)}+c \)

Explanation:
1. Split: \( \frac{1}{\cos^2x}-\frac{1}{\sin^2x}=\sec^2x-\operatorname{cosec}^2x \).
2. Integrating gives \( \tan x+\cot x+c \).

Teacher's Note:
1. Divide each term of the numerator by the denominator.
2. Then use the two standard integrals sec squared and cosec squared.

Answer: (c) \( \log{\left(\sin{\left(x + \frac{\pi}{4} \right)} \right)}+c \)

Explanation:
1. \( \frac{1-\tan x}{1+\tan x}=\tan\left(\frac\pi4-x\right) \).
2. \( \int\tan\left(\frac\pi4-x\right)dx=\log\cos\left(\frac\pi4-x\right)+c=\log\sin\left(\frac\pi4+x\right)+c \), because \( \cos\left(\frac\pi4-x\right)=\sin\left(\frac\pi4+x\right) \).

Teacher's Note:
1. Recognise the tangent subtraction formula with tan pi/4 = 1.
2. The inner function has derivative minus 1.

Answer: (d) \( - \log{\left(\tan{\left(x \right)} + \sec{\left(x \right)} \right)} + 2 \sin{\left(x \right)}+c \)

Explanation:
1. \( \frac{\cos2x}{\cos x}=\frac{2\cos^2x-1}{\cos x}=2\cos x-\sec x \).
2. Integrating: \( 2\sin x-\log(\sec x+\tan x)+c \).

Teacher's Note:
1. Use cos 2x = 2 cos squared x minus 1.
2. Do not forget the minus sign before the integral of sec x.

Answer: (a) \( \frac{2 x^{\frac{7}{2}}}{7} + \frac{6 x^{\frac{5}{2}}}{5} + 2 x^{\frac{3}{2}} + 2 \sqrt{x}+c \)

Explanation:
1. Expand: \( (1+x)^3=1+3x+3x^2+x^3 \), so the integrand is \( x^{-1/2}+3x^{1/2}+3x^{3/2}+x^{5/2} \).
2. Integrating each term gives \( 2x^{1/2}+2x^{3/2}+\frac65x^{5/2}+\frac27x^{7/2}+c \).

Teacher's Note:
1. Expand the numerator and divide by root x term by term.
2. Check each coefficient by dividing by the new exponent.

Answer: (c) \( \frac{\log{\left(\frac{3 x - 2}{3 x + 2} \right)}}{12}+c \)

Explanation:
1. \( \frac{1}{9x^2-4}=\frac19\cdot\frac{1}{x^2-(2/3)^2} \) and \( \int\frac{dx}{x^2-k^2}=\frac{1}{2k}\log\left|\frac{x-k}{x+k}\right| \) with \( k=\frac23 \).
2. This gives \( \frac19\cdot\frac34\log\left|\frac{x-2/3}{x+2/3}\right|=\frac{1}{12}\log\left|\frac{3x-2}{3x+2}\right|+c \).

Teacher's Note:
1. Factor 9 out so the denominator has the standard form x squared minus k squared.
2. The fraction inside the log has x minus k on top.

Answer: (c) \( \frac{\log{\left(\frac{a + x}{a - x} \right)}}{2 a}+c \)

Explanation:
1. Use partial fractions: \( \frac{1}{a^2-x^2}=\frac{1}{2a}\left(\frac{1}{a+x}+\frac{1}{a-x}\right) \).
2. Integrating gives \( \frac{1}{2a}\left[\log(a+x)-\log(a-x)\right]=\frac{1}{2a}\log\frac{a+x}{a-x}+c \) for \( |x|<a \).

Teacher's Note:
1. Learn the pair: log of (x - a)/(x + a) for x squared minus a squared, and the reverse for a squared minus x squared.
2. Check the fraction by differentiating.

Answer: (b) \( \frac{a^{2} \log{\left(x + \sqrt{a^{2} + x^{2}} \right)}}{2} + \frac{x \sqrt{a^{2} + x^{2}}}{2}+c \)

Explanation:
1. This is a standard result: \( \int\sqrt{x^2+a^2}dx=\frac x2\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left(x+\sqrt{x^2+a^2}\right)+C \).
2. It can be derived by integration by parts or by putting \( x=a\tan\theta \).

Teacher's Note:
1. Memorise the three standard root integrals together with their signs.
2. For x squared plus a squared both terms are added.

Answer: (c) \( \frac{\tan^{-1}{\left(\frac{2 x}{3} \right)}}{6}+c \)

Explanation:
1. \( \int\frac{dx}{4x^2+9}=\frac14\int\frac{dx}{x^2+(3/2)^2}=\frac14\cdot\frac{2}{3}\tan^{-1}\frac{2x}{3} \).
2. So the result is \( \frac16\tan^{-1}\frac{2x}{3}+c \).

Teacher's Note:
1. Factor out the coefficient of x squared first.
2. Then use the 1 over (x squared plus k squared) formula with k = 3/2.

Answer: (d) \( \frac{2 x^{\frac{3}{2}}}{3} + \frac{4}{3}+c \)

Explanation:
1. \( f(x)=\int x^{1/2}dx=\frac23x^{3/2}+C \).
2. \( f(1)=\frac23+C=2 \), so \( C=\frac43 \) and \( f(x)=\frac23\left(x\sqrt x+2\right) \).

Teacher's Note:
1. Integrate first and then use the given point to find C.
2. Write the final answer in the form given in the options.

Answer: (b) \( - \frac{x^{2}}{2}+c \)

Explanation:
1. For \( x<0 \), \( |x|=-x \).
2. \( \int(-x)dx=-\frac{x^2}{2}+c \).

Teacher's Note:
1. Remove the modulus using the sign of x on the given interval.
2. The derivative of minus x squared over 2 is minus x, which is positive when x is negative.

Answer: (a) \( - x - 2 \cot{\left(x \right)} + 2 \csc{\left(x \right)}+c \)

Explanation:
1. Multiply numerator and denominator by \( \sin\theta \): \( \frac{1-\cos\theta}{1+\cos\theta}=\tan^2\frac\theta2=\sec^2\frac\theta2-1 \).
2. Integrating: \( 2\tan\frac\theta2-\theta+c=2\operatorname{cosec}\theta-2\cot\theta-\theta+c \), since \( \tan\frac\theta2=\operatorname{cosec}\theta-\cot\theta \).

Teacher's Note:
1. The half-angle identity converts the fraction into tan squared of half the angle.
2. Use tan of half angle equals cosec minus cot to match the options.

Answer: (c) \( \frac{a^{x}}{\log{\left(a \right)}} + \frac{x^{a + 1}}{a + 1}+c \)

Explanation:
1. \( e^{a\log x}=x^a \) and \( e^{x\log a}=a^x \).
2. \( \int x^adx=\frac{x^{a+1}}{a+1} \) and \( \int a^xdx=\frac{a^x}{\log a} \), so the sum is \( \frac{x^{a+1}}{a+1}+\frac{a^x}{\log a}+c \).

Teacher's Note:
1. Simplify each exponential-log expression first.
2. Here a is a constant, so the power rule and the exponential rule are both used.

Answer: (b) \( \frac{\log{\left(\tan{\left(6 x \right)} + \sec{\left(6 x \right)} \right)}}{6}+c \)

Explanation:
1. With \( \theta=2x \), \( 4\cos^3\theta-3\cos\theta=\cos3\theta=\cos6x \).
2. \( \int\sec6x\,dx=\frac16\log(\sec6x+\tan6x)+c \).

Teacher's Note:
1. Recognise the triple angle formula for cosine.
2. Divide by 6 because the angle is 6x.

Answer: (c) \( \tan{\left(x \right)} - \cot{\left(x \right)}+c \)

Explanation:
1. Write \( 1=\sin^2x+\cos^2x \): the integrand is \( \sec^2x+\operatorname{cosec}^2x \).
2. Integrating gives \( \tan x-\cot x+c \).

Teacher's Note:
1. Replace 1 by sin squared plus cos squared in the numerator.
2. Integral of cosec squared is minus cot, so the sign becomes minus.

Answer: (a) \( 2 x \cos{\left(a \right)} + 2 \sin{\left(x \right)}+c \)

Explanation:
1. \( \cos2x-\cos2\alpha=2(\cos^2x-\cos^2\alpha)=2(\cos x-\cos\alpha)(\cos x+\cos\alpha) \).
2. The integrand reduces to \( 2(\cos x+\cos\alpha) \) and the integral is \( 2(\sin x+x\cos\alpha)+c \).

Teacher's Note:
1. Express cos 2x with the square formula and factor the difference of squares.
2. Treat alpha as a constant.