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Chapter 11: Problem 916

If \(\int_{x \operatorname{cosec}^{2} x d x}=P \cdot x \cot x+Q \log |\sin x|+C\) then \(P+Q=\) (a) 1 (b) 2 (c) 0 (d) \(-1\)

Short Answer

Expert verified
The sum of P and Q is 0. The correct answer is (c) 0.

Step by step solution

01

Write down the given integral

The given integral is \(\int x\operatorname{cosec}^2x dx\).
02

Use the integration by parts method

To integrate by parts, we need to choose a part that we can easily differentiate, and a part that we can integrate. We choose \(u = x\) and \(dv = \operatorname{cosec}^2{x} dx\). Therefore, we have: \(du = dx\) \(v = \int \operatorname{cosec}^2{x} dx = -\cot x\) Now we apply the integration by parts formula: \(\int u dv = uv - \int v du\).
03

Apply the formula

Using the integration by parts formula, we get: \(\int x\operatorname{cosec}^2x dx = -x\cot x + \int (\cot x) dx\)
04

Integrate the remaining part

Now, we need to integrate \(\int \cot x dx\). Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\frac{d(\sin x)}{dx} = \cos x\). So, we can write \(\int \cot x dx\) as \(\int \frac{\cos x}{\sin x} dx\), and then we can make the substitution \(u=\sin x, du=\cos x dx\). Thus, the remaining integral becomes: \(\int \cot x dx = \int \frac{du}{u} = \log |u| + C = \log|\sin x| + C\)
05

Combine the results

Putting the results obtained in steps 3 and 4 together: \(\int x\operatorname{cosec}^2x dx = -x\cot x + \log|\sin x| + C\) Comparing this to the given result \(Px\cot x + Q\log|\sin x| + C\), we can see that \(P = -1\) and \(Q = 1\).
06

Find the sum of P and Q

Since we have determined the values of P and Q, we can now calculate the sum of P and Q: \(P + Q = -1 + 1 = 0\) The correct answer is (c) 0.

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Most popular questions from this chapter

\(\int \mathrm{e}^{\mathrm{x}}\left[(1-\mathrm{x}) /\left(1+\mathrm{x}^{2}\right)\right]^{2} \mathrm{dx}=\underline{\mathrm{c}}\) (a) \(e^{x}\left(1+x^{2}\right)\) (b) \(\left[\mathrm{e}^{\mathrm{x}} /\left(1+\mathrm{x}^{2}\right)\right]\) (c) \(\mathrm{e}^{\mathrm{x}}\left[(1-\mathrm{x}) /\left(1+\mathrm{x}^{2}\right)\right]\) (d) \(e^{x}\left(1-x^{2}\right)\)

if \(\int\left[\left(4 \mathrm{e}^{\mathrm{x}}+6 \mathrm{e}^{-\mathrm{x}}\right) /\left(9 \mathrm{e}^{\mathrm{x}}-4 \mathrm{e}^{-\mathrm{x}}\right)\right] \mathrm{dx}=\mathrm{Ax}+\mathrm{B} \log \left(9 \mathrm{e}^{2 \mathrm{x}}-4\right)+\mathrm{c}\) then \(\mathrm{A}, \mathrm{B}=\) (a) \((3 / 2),[(-35) /(36)]\) (b) \(-(3 / 2),[(-35) /(36)]\) (c) \(-(3 / 2),(35 / 36)\) (d) \((3 / 2),(35 / 36)\)

\(\int \operatorname{cosec}^{3} x d x=\ldots+c\) (a) \(-(1 / 2) \operatorname{cosec} x \cot x+(1 / 2) \log |(\operatorname{cosec} x-\cot x)|\) (b) \(-(1 / 2) \operatorname{cosec} x \cot x\) (c) \((1 / 2) \operatorname{cosec} x \cot x+(1 / 2) \log |(\operatorname{cosec} x-\cot x)|\) (d) \((1 / 2) \operatorname{cosec} x \cot x-(1 / 2) \log \mid(\operatorname{cosec} x-\cot x)\)

\(\int\left[\left(\cos ^{8} x-\sin ^{8} x\right) /\left(1-2 \sin ^{2} x \cos ^{2} x\right)\right] d x=\underline{ }+c\) (a) \(-[(\cos 2 \mathrm{x}) / 2]\) (b) \(\overline{-[(\sin 2} x) / 2]\) (c) \([(\cos 2 \mathrm{x}) / 2]\) (d) \([(\sin 2 \mathrm{x}) / 2]\)

If \(\int[(2 \sin x+\cos x) /(7 \sin x-5 \cos x)] d x\) \(=a x+b \log |7 \sin x-5 \cos x|+c\) then \(a-b=\) (a) \((4 / 37)\) (b) \(-(4 / 37)\) (c) \((8 / 37)\) (d) \(-(8 / 37)\)
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