JEE Mathematics Indefinite Integration MCQs Set 06

Answer: (a) \( x - \log{\left(\sqrt{1 - e^{2 x}} + 1 \right)}+c \)

Explanation:
1. Put \( t=\sqrt{1-e^{2x}} \); then \( e^{2x}=1-t^2 \) and \( dx=-\frac{t\,dt}{1-t^2} \).
2. Since \( \frac{1+t}{1-t}=\frac{(1+t)^2}{1-t^2}=\frac{(1+t)^2}{e^{2x}} \), we get \( -\frac12\log\frac{1+t}{1-t}=-\log(1+t)+x+c \), i.e. \( x-\log\left(1+\sqrt{1-e^{2x}}\right)+c \).

Teacher's Note:
1. Rationalising substitution t = root of (1 - e to the 2x).
2. Verify by differentiating the final answer.

Answer: (a) \( \log{\left(\sqrt{\tan^{2}{\left(x \right)} + 4} + \tan{\left(x \right)} \right)}+c \)

Explanation:
1. Put \( t=\tan x \), \( dt=\sec^2x\,dx \).
2. \( \int\frac{dt}{\sqrt{t^2+4}}=\log\left(t+\sqrt{t^2+4}\right)+c \).

Teacher's Note:
1. Use the standard form 1 over root of t squared plus a squared.
2. Always substitute the tangent when sec squared appears as a factor.

Answer: (b) \( \frac{\sqrt{4 - \sin^{2}{\left(x \right)}} \sin{\left(x \right)}}{2} + 2 \sin^{-1}{\left(\frac{\sin{\left(x \right)}}{2} \right)}+c \)

Explanation:
1. Put \( t=\sin x \), \( dt=\cos x\,dx \): \( \int\sqrt{4-t^2}\,dt \).
2. Using \( \int\sqrt{a^2-t^2}dt=\frac t2\sqrt{a^2-t^2}+\frac{a^2}{2}\sin^{-1}\frac ta \) with \( a=2 \), the result is \( \frac12\sin x\sqrt{4-\sin^2x}+2\sin^{-1}\left(\frac12\sin x\right)+c \).

Teacher's Note:
1. Memorise the standard integral of root of a squared minus t squared.
2. Here a squared over 2 equals 2.

Answer: (a) \( \frac{\sin^{-1}{\left(\frac{4 x^{3}}{3} \right)}}{4}+c \)

Explanation:
1. Put \( t=x^3 \), \( dt=3x^2dx \): \( \int\frac{dt}{\sqrt{9-16t^2}}=\frac14\int\frac{d(4t)}{\sqrt{9-(4t)^2}} \).
2. \( \frac14\sin^{-1}\frac{4t}{3}=\frac14\sin^{-1}\frac{4x^3}{3}+c \).

Teacher's Note:
1. Write 16 t squared as (4t) squared.
2. Divide by 4 for the inner derivative.

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

Explanation:
1. Put \( t=x^2 \), \( x\,dx=\frac{dt}{2} \).
2. \( \frac12\int\frac{dt}{\sqrt{4-t^2}}=\frac12\sin^{-1}\frac t2=\frac12\sin^{-1}\frac{x^2}{2}+c \).

Teacher's Note:
1. Substitute x squared and keep the factor 1/2.
2. Option (b) has the wrong inverse function.

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

Explanation:
1. Put \( t=a^x \), \( dt=a^x\log a\,dx \).
2. \( \frac{1}{\log a}\int\frac{dt}{\sqrt{1-t^2}}=\frac{1}{\log a}\sin^{-1}a^x+c \).

Teacher's Note:
1. Divide by log a because of the derivative of a to the x.
2. The integral is arcsine, not arccosine.

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

Explanation:
1. Put \( t=\cos x \), \( \sin x\,dx=-dt \): \( -\int\frac{dt}{3+4t^2}=-\frac14\int\frac{dt}{t^2+3/4} \).
2. \( -\frac14\cdot\frac{2}{\sqrt3}\tan^{-1}\frac{2t}{\sqrt3}=-\frac{1}{2\sqrt3}\tan^{-1}\left(\frac{2\cos x}{\sqrt3}\right)+c \).

Teacher's Note:
1. Factor the coefficient of t squared out before applying the arctan formula.
2. The minus sign comes from the derivative of cos x.

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

Explanation:
1. Put \( t=a^2+b^2\sin^2x \); then \( dt=b^2\sin2x\,dx \).
2. \( \frac{1}{b^2}\int\frac{dt}{t}=\frac{1}{b^2}\log(a^2+b^2\sin^2x)+c \).

Teacher's Note:
1. The derivative of sin squared x is sin 2x.
2. The constant b squared must be divided out.

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

Explanation:
1. Let \( t=a\cos^2x+b\sin^2x \); then \( dt=2(b-a)\sin x\cos x\,dx \).
2. \( \frac{1}{2(b-a)}\log t+c \).

Teacher's Note:
1. Differentiate the denominator first and compare with the numerator.
2. The derivative contains the factor b minus a.

Answer: (c) \( 2 \sqrt{\log{\left(x \right)} + 1}+c \)

Explanation:
1. Put \( t=1+\log x \), \( dt=\frac{dx}{x} \).
2. \( \int t^{-1/2}dt=2\sqrt t=2\sqrt{1+\log x}+c \).

Teacher's Note:
1. The exponent minus 1/2 becomes 1/2 after integration, with a factor 2.
2. Do not confuse with the integral of a positive power.

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

Explanation:
1. The integrand is \( \frac{1}{x(1+\log x)} \). Put \( t=1+\log x \), \( dt=\frac{dx}{x} \).
2. \( \int\frac{dt}{t}=\log(1+\log x)+c \).

Teacher's Note:
1. Factor x from the denominator.
2. The inner function 1 + log x has derivative 1/x.

Answer: (b) \( \log{\left(\sin^{2}{\left(x \right)} + 1 \right)}+c \)

Explanation:
1. The derivative of \( 1+\sin^2x \) is \( \sin2x \).
2. So the integral is \( \log(1+\sin^2x)+c \).

Teacher's Note:
1. The numerator is exactly the derivative of the denominator.
2. No extra constant is needed.

Answer: (c) \( \log{\left(\tan{\left(x \right)} + 1 \right)}+c \)

Explanation:
1. The derivative of \( 1+\tan x \) is \( \sec^2x \).
2. So the integral is \( \log(1+\tan x)+c \).

Teacher's Note:
1. Numerator equals the derivative of the denominator.
2. Result is log of the denominator.

Answer: (b) \( - \log{\left(\cot{\left(x \right)} + 1 \right)}+c \)

Explanation:
1. The derivative of \( 1+\cot x \) is \( -\operatorname{cosec}^2x \).
2. So the integral is \( -\log(1+\cot x)+c \).

Teacher's Note:
1. The minus sign appears because the derivative of cot x is negative.
2. Check by differentiating.

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

Explanation:
1. Put \( t=\sqrt x \), \( \frac{dx}{\sqrt x}=2dt \).
2. \( 2\int\sin t\,dt=-2\cos\sqrt x+c \).

Teacher's Note:
1. The derivative of root x is 1 over 2 root x, so a factor 2 appears.
2. Then apply the integral of sine.

Answer: (c) \( \sqrt{x^{2} + 1} + \log{\left(x + \sqrt{x^{2} + 1} \right)}+c \)

Explanation:
1. Split: \( \int\frac{x\,dx}{\sqrt{1+x^2}}+\int\frac{dx}{\sqrt{1+x^2}} \).
2. The first is \( \sqrt{1+x^2} \) and the second is \( \log\left(x+\sqrt{1+x^2}\right) \).

Teacher's Note:
1. Split the numerator into two separate integrals.
2. The second is a standard log form.

Answer: (a) \( \frac{\log{\left(3^{x} + \sqrt{9^{x} - 1} \right)}}{\log{\left(3 \right)}}+c \)

Explanation:
1. Put \( t=3^x \), \( dt=3^x\log3\,dx \); the integral is \( \frac{1}{\log3}\int\frac{dt}{\sqrt{t^2-1}} \).
2. \( \frac{1}{\log3}\log\left|t+\sqrt{t^2-1}\right|=\frac{1}{\log3}\log\left|3^x+\sqrt{9^x-1}\right|+c \).

Teacher's Note:
1. Write 9 to the x as (3 to the x) squared.
2. Divide by log 3, not by log 9.

Answer: (b) \( \log{\left(10^{x} + x^{10} \right)}+c \)

Explanation:
1. The derivative of \( 10^x+x^{10} \) is \( 10^x\log_e10+10x^9 \).
2. So the integral is \( \log\left(10^x+x^{10}\right)+c \).

Teacher's Note:
1. Compare the numerator with the derivative of the denominator.
2. Result is log of the denominator.

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

Explanation:
1. Write \( \sin x=\sin\left[(x-\alpha)+\alpha\right]=\sin(x-\alpha)\cos\alpha+\cos(x-\alpha)\sin\alpha \).
2. The integrand is \( \cos\alpha+\sin\alpha\cot(x-\alpha) \), so the integral is \( x\cos\alpha+\sin\alpha\,\log\sin(x-\alpha)+c \).

Teacher's Note:
1. Write x as (x minus alpha) plus alpha.
2. Then split into a constant and a cotangent term.

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

Explanation:
1. Put \( t=\tan^{-1}x^2 \); then \( dt=\frac{2x}{1+x^4}dx \).
2. \( \int t\,dt=\frac{t^2}{2}=\frac12\left[\tan^{-1}x^2\right]^2+c \).

Teacher's Note:
1. The factor 2x over 1 + x to the 4 is the derivative of tan inverse of x squared.
2. The remaining integral is t.

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

Explanation:
1. For \( |x|<1 \), \( \tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x \).
2. By parts, \( \int\tan^{-1}x\,dx=x\tan^{-1}x-\frac12\log(1+x^2) \), so the result is \( 2x\tan^{-1}x-\log(1+x^2)+c \).

Teacher's Note:
1. Use the double angle formula for inverse tangent in the correct range.
2. The integral of tan inverse x is found by parts.

Answer: (d) \( 3 x \sin^{-1}{\left(x \right)} + 3 \sqrt{1 - x^{2}}+c \)

Explanation:
1. Put \( x=\sin\theta \): \( \sin^{-1}(3\sin\theta-4\sin^3\theta)=\sin^{-1}(\sin3\theta)=3\theta \) for \( |x|\le\frac12 \).
2. So the integrand is \( 3\sin^{-1}x \) and \( \int\sin^{-1}x\,dx=x\sin^{-1}x+\sqrt{1-x^2} \). The result is \( 3\left[x\sin^{-1}x+\sqrt{1-x^2}\right]+c \).

Teacher's Note:
1. Recognise the triple angle formula for sine.
2. The result is valid on the principal range of the inverse function.

Answer: (d) \( \sin^{-1}{\left(2 x \right)}+c \)

Explanation:
1. \( \int\frac{dx}{\sqrt{1-(2x)^2}}=\frac12\sin^{-1}(2x) \).
2. Multiplying by 2 gives \( \sin^{-1}(2x)+c \).

Teacher's Note:
1. The inner derivative 2 cancels with the numerator 2.
2. Arcsine, not arccosine, is the antiderivative of this form.

Answer: (b) \( \log(\log\operatorname{cosec}x)+c \)

Explanation:
1. Put \( t=\log\sin x \); then \( dt=\cot x\,dx \).
2. \( \int\frac{dt}{t}=\log|t|=\log|\log\sin x|=\log\left(\log\operatorname{cosec}x\right)+c \), since \( \log\sin x<0 \) for \( 0<\sin x<1 \).

Teacher's Note:
1. The numerator cot x is the derivative of log sin x.
2. Since log sin x is negative, the real form is log of log cosec x.

Answer: (a) \( \dfrac12[f(x)]^2 \)

Explanation:
1. Differentiating \( \int f\,dx=f \) gives \( f(x)=f'(x) \), so \( f(x)=Ce^x \).
2. \( \int[f(x)]^2dx=C^2\int e^{2x}dx=\frac{C^2e^{2x}}{2}=\frac12[f(x)]^2+k \).

Teacher's Note:
1. The condition f dash equals f forces an exponential function.
2. The square of an exponential integrates with a factor 1/2.