Short Quiz - Chapter 7 Class 12 Integrals

Correct option: D

Answer: $$ \int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x $$


Use:

$$ \sin 2 x=2 \sin x \cos x $$


So,

$$ 1-\sin 2 x=1-2 \sin x \cos x $$


But:

$$ 1=\sin ^2 x+\cos ^2 x $$


Therefore,

$$ \begin{aligned} 1-\sin 2 x= & \sin ^2 x+\cos ^2 x-2 \sin x \cos x \\ & =(\sin x-\cos x)^2 \end{aligned} $$


So,

$$ \sqrt{1-\sin 2 x}=|\sin x-\cos x| $$


Now, split the integral at:

$$ x=\frac{\pi}{4} $$

because at \(x=\frac{\pi}{4}\),

$$ \sin x=\cos x $$


From 0 to \(\frac{\pi}{4}\) :

$$ \cos x>\sin x $$


So,

$$ |\sin x-\cos x|=\cos x-\sin x $$


From \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\) :

$$ \sin x>\cos x $$


So,

$$ |\sin x-\cos x|=\sin x-\cos x $$


Therefore,

$$ \int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x=\int_0^{\pi / 4}(\cos x-\sin x) d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x $$


First integral:

$$ \begin{gathered} \int_0^{\pi / 4}(\cos x-\sin x) d x=[\sin x+\cos x]_0^{\pi / 4} \\ =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1) \\ =\sqrt{2}-1 \end{gathered} $$


Second integral:

$$ \begin{gathered} \int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x=[-\cos x-\sin x]_{\pi / 4}^{\pi / 2} \\ =(0-1)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) \\ =-1+\sqrt{2} \\ =\sqrt{2}-1 \end{gathered} $$


Total:

$$ \begin{gathered} (\sqrt{2}-1)+(\sqrt{2}-1) \\ =2(\sqrt{2}-1) \\ 2(\sqrt{2}-1) \end{gathered} $$


Answer: (D)

Correct option: C

Answer: Solution (C) is the correct answer, since differentiating both sides, we have

$$ \frac{3 e^x-5 e^{-x}}{4 e^x+5 e^{-x}}=a+b \frac{\left(4 e^x-5 e^{-x}\right)}{4 e^x+5 e^{-x}}, $$

giving \(3 e^x-5 e^{-x}=a\left(4 e^x+5 e^{-x}\right)+b\left(4 e^x-5 e^{-x}\right)\). Comparing coefficients on both sides, we get \(3=4 a+4 b\) and \(-5=5 a-5 b\). This verifies \(a=\frac{-1}{8}, b=\frac{7}{8}\).

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Correct option: B

Answer: Given:
- The integral \(\int_{-1}^1(1-|x|) d x\)

To Find: An equivalent expression from the options.
Step-by-Step Reasoning:
1. Check for symmetry (Even/Odd Function):

Let \(f(x)=1-|x|\).
Test \(f(-x)\) :

$$ f(-x)=1-|-x|=1-|x|=f(x) $$


Since \(f(-x)=f(x)\), the function is an even function.
2. Apply the property of definite integrals for even functions:

For an even function, \(\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\), which by symmetry is also equal to \(2 \int_{-a}^0 f(x) d x\).
Let's use the negative half of the interval:

$$ \int_{-1}^1(1-|x|) d x=2 \int_{-1}^0(1-|x|) d x $$

3. Simplify the absolute value on the chosen interval:

On the interval \([-1,0]\), the value of \(x\) is negative (or zero).
Therefore, by definition of absolute value, \(|x|=-x\).
Substitute this into our integrand:

$$ 1-|x|=1-(-x)=1+x $$

4. Finalize the expression:

$$ 2 \int_{-1}^0(1+x) d x $$


Answer:
The correct option is B) \(2 \int_{-1}^0(1+x) d x\).

Correct option: A

Answer: $$ \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x $$


Notice that this looks like the derivative of:

$$ \frac{e^x}{1+x^2} $$


Differentiate it:

$$ \begin{gathered} \frac{d}{d x}\left(\frac{e^x}{1+x^2}\right)=\frac{e^x\left(1+x^2\right)-e^x(2 x)}{\left(1+x^2\right)^2} \\ =\frac{e^x\left(1+x^2-2 x\right)}{\left(1+x^2\right)^2} \\ =\frac{e^x\left(x^2-2 x+1\right)}{\left(1+x^2\right)^2} \\ =e^x\left(\frac{x-1}{1+x^2}\right)^2 \end{gathered} $$


Since:

$$ (x-1)^2=(1-x)^2 $$


So,

$$ e^x\left(\frac{x-1}{1+x^2}\right)^2=e^x\left(\frac{1-x}{1+x^2}\right)^2 $$


Hence,

$$ \frac{e^x}{1+x^2}+C $$


Answer: (A)