JEE Mathematics Indefinite Integration MCQs Set 03

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

Explanation:
1. The standard result is \( \int \sec x\,dx=\log|\sec x+\tan x|+c=\log\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)+c \).
2. Option (a) has the wrong sign inside the tangent and option (b) has the wrong sign between sec x and tan x.
3. Option (c) is not an antiderivative either, so (d) is correct.

Teacher's Note:
1. Memorise both forms of the integral of sec x: the log of sec x plus tan x and the log tan of pi/4 plus x/2.
2. Differentiating a candidate answer is a quick way to confirm it.

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

Explanation:
1. \( \int \sin x\,dx=-\cos x \), and the constant 5 comes out of the integral.
2. So the result is \( -5\cos x+c \).

Teacher's Note:
1. A constant factor can always be taken outside an integral.
2. Do not forget the minus sign in the integral of sin x.

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

Explanation:
1. Expand: \( (\sec x+\tan x)^2=\sec^2x+\tan^2x+2\sec x\tan x=2\sec^2x-1+2\sec x\tan x \).
2. Integrate term by term: \( 2\tan x-x+2\sec x \).
3. This equals \( 2(\sec x+\tan x)-x+c \).

Teacher's Note:
1. Rewrite tan squared as sec squared minus 1 before integrating.
2. Option (b) is the wrong power rule, because the derivative of sec x + tan x is not 1.

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

Explanation:
1. Since \( \frac{d}{dx}(\cot x)=-\operatorname{cosec}^2x \), the integral is \( -\cot x+c \).

Teacher's Note:
1. The integral of cosec squared x is minus cot x.
2. Learn the six basic trigonometric integrals together with their derivatives.

Answer: (b) \( \sec{\left(x \right)}+c \)

Explanation:
1. Since \( \frac{d}{dx}(\sec x)=\sec x\tan x \), the integral is \( \sec x+c \).

Teacher's Note:
1. Derivative of sec x is sec x tan x, so the integral reverses it.
2. Option (a) wrongly adds tan x.

Answer: (c) \( x+c \)

Explanation:
1. \( 1+\sin2x=(\sin x+\cos x)^2 \), so the integrand is \( \frac{\sin x+\cos x}{|\sin x+\cos x|} \).
2. Where \( \sin x+\cos x>0 \) the integrand is 1, so the integral is \( x+c \).

Teacher's Note:
1. Recognise 1 + sin 2x as a perfect square.
2. The result holds on the interval where sin x + cos x is positive.

Answer: (b) \( - \frac{2 \cos{\left(3 x \right)}}{3} - 3 \cot{\left(x \right)}+c \)

Explanation:
1. \( \int3\operatorname{cosec}^2x\,dx=-3\cot x \) and \( \int2\sin3x\,dx=-\frac{2}{3}\cos3x \).
2. The result is \( -\left(3\cot x+\frac23\cos3x\right)+c \).

Teacher's Note:
1. Dividing by the coefficient of x inside the trig function is needed for sin 3x.
2. Both parts give negative signs.

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

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

Teacher's Note:
1. Split the fraction and use cot squared = cosec squared minus 1.
2. Always reduce to standard integrals first.

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

Explanation:
1. \( \int\cot x\,dx=\int\frac{\cos x}{\sin x}dx=\log|\sin x|+c \).

Teacher's Note:
1. Integral of f dash over f is log of f.
2. The integral of tan x is log sec x, not log of cot x.

Answer: (b) \( - \frac{1}{x - 5}+c \)

Explanation:
1. \( \int(x-5)^{-2}dx=\frac{(x-5)^{-1}}{-1}=-\frac{1}{x-5}+c \).

Teacher's Note:
1. Use the power rule for a linear base.
2. The exponent increases by 1 and you divide by the new exponent.

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

Explanation:
1. \( \frac{x^2}{x^2+4}=1-\frac{4}{x^2+4} \).
2. Since \( \int\frac{dx}{x^2+4}=\frac12\tan^{-1}\frac x2 \), the result is \( x-2\tan^{-1}\frac x2+c \).

Teacher's Note:
1. When degrees of numerator and denominator are equal, divide first.
2. The standard form is 1 over x squared plus a squared.

Answer: (c) \( \frac{\log{\left(\tan{\left(x^{3} \right)} + \sec{\left(x^{3} \right)} \right)}}{3}+c \)

Explanation:
1. Put \( t=x^3 \), so \( x^2dx=\frac{dt}{3} \).
2. \( \frac13\int\sec t\,dt=\frac13\log(\sec t+\tan t)+c \).

Teacher's Note:
1. Substitute the inner function whose derivative is present.
2. Do not forget the factor 1/3 from dt.

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

Explanation:
1. \( \cos2x-1=-2\sin^2x \) and \( \cos2x+1=2\cos^2x \), so the integrand is \( -\tan^2x \).
2. \( -\int(\sec^2x-1)dx=-\tan x+x+c \).

Teacher's Note:
1. Use half-angle identities for the numerator and denominator.
2. Then convert tan squared to sec squared minus 1.

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

Explanation:
1. For \( 0\le x\le\pi \), \( \sin^{-1}(\cos x)=\frac{\pi}{2}-x \).
2. Integrating gives \( \frac{\pi x}{2}-\frac{x^2}{2}+c=\frac{\pi x-x^2}{2}+c \).

Teacher's Note:
1. Write cos x as sin of pi/2 minus x and keep the angle in the principal range.
2. The principal range of inverse sine is minus pi/2 to pi/2.

Answer: (a) \( \frac{\pi x}{2}+c \)

Explanation:
1. \( \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} \) for \( -1\le x\le1 \).
2. So the integral is \( \frac{\pi}{2}x+c \).

Teacher's Note:
1. Complementary inverse functions add to a constant.
2. Do not integrate the two inverse functions separately.

Answer: (a) \( \frac{x^{52} \left(\cot^{-1}{\left(x \right)} + \tan^{-1}{\left(x \right)}\right)}{52}+c \)

Explanation:
1. \( \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} \) for \( x>0 \).
2. \( \frac\pi2\int x^{51}dx=\frac{\pi}{2}\cdot\frac{x^{52}}{52}=\frac{x^{52}}{52}(\tan^{-1}x+\cot^{-1}x)+c \).

Teacher's Note:
1. Use the constant sum of tan inverse and cot inverse.
2. Then apply the power rule.

Answer: (a) \( - \frac{1}{3 x^{3}}+c \)

Explanation:
1. \( \int x^{-4}dx=\frac{x^{-3}}{-3}+c=-\frac{1}{3x^3}+c \).

Teacher's Note:
1. Add 1 to the exponent -4 to get -3, then divide by -3.
2. Check the sign of the exponent carefully.

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

Explanation:
1. \( \frac{d}{dx}\left(\frac{a^x}{\log a}\right)=a^x \), so \( \int a^xdx=\frac{a^x}{\log a}+C \).

Teacher's Note:
1. The integral of an exponential with base a divides by log a.
2. Multiplying by log a gives the derivative, not the integral.

Answer: (d) \( \frac{a^{x+1}}{x+1}+C \)

Explanation:
1. Here the variable of integration is a and x is a constant, so it is a power rule: \( \int a^xda=\frac{a^{x+1}}{x+1}+C \).
2. Options (a) and (b) integrate with respect to x and option (c) has the wrong exponent.

Teacher's Note:
1. Always read which letter is the variable of integration.
2. Here the power rule applies, not the exponential rule.

Answer: (a) \( \frac{13^{x}}{\log{\left(13 \right)}}+c \)

Explanation:
1. \( \int a^xdx=\frac{a^x}{\log a} \) with \( a=13 \).
2. So the integral is \( \frac{13^x}{\log13}+C \).

Teacher's Note:
1. Apply the exponential rule with base 13.
2. Do not use the power rule on a variable exponent.

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

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

Teacher's Note:
1. Combine the two exponentials into a single exponential with base a e.
2. Then apply the exponential rule.

Answer: (b) \( \frac{a^{3 x + 3}}{3 \log{\left(a \right)}}+c \)

Explanation:
1. Use \( \int a^{px+q}dx=\frac{a^{px+q}}{p\log a} \) with \( p=3 \).
2. The result is \( \frac{a^{3x+3}}{3\log a}+c \).

Teacher's Note:
1. Divide by the coefficient of x as well as by log a.
2. Check by differentiating: the chain rule gives back 3 log a.

Answer: (b) \( -\cos x+c \)

Explanation:
1. \( e^{\log(\sin x)}=\sin x \).
2. \( \int\sin x\,dx=-\cos x+c \).

Teacher's Note:
1. e to the power log of f is just f.
2. Always simplify before integrating.

Answer: (a) \( \frac{x^{m + 1}}{m + 1}+c \)

Explanation:
1. \( e^{m\log x}=x^m \).
2. \( \int x^mdx=\frac{x^{m+1}}{m+1}+k \) for \( m\ne-1 \).

Teacher's Note:
1. Use e to the power m log x equals x to the power m.
2. Then use the power rule.

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

Explanation:
1. The integrand is \( \frac{x^5-x^4}{x^3-x^2}=\frac{x^4(x-1)}{x^2(x-1)}=x^2 \).
2. \( \int x^2dx=\frac{x^3}{3}+c \).

Teacher's Note:
1. Convert each exponential of a log to a power of x.
2. Cancel the common factor x minus 1.