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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{dx} / \sqrt{(x-4)(x-7)}]=+c \quad(4

\(\int \sqrt{(1+\operatorname{cosec} x d x)}=\ldots\) (a) \(2 \sin ^{-1}(\sqrt{\cos } \mathrm{x})\) (b) \(2 \cos ^{-1}(\sqrt{\sin } \mathrm{x})\) (c) \(\sin ^{-1}(2 \sin x-1)\) (d) \(2 \cos ^{-1}(\sqrt{\cos x})\)

\(\int e^{x}\left[\left(x^{3}-x-2\right) /\left(x^{2}+1\right)^{2}\right] d x=\) (a) \(e^{x}\left[(2 x-1) /\left(x^{2}+1\right)\right]\) (b) \(e^{x}\left[(x+1) /\left(x^{2}+1\right)\right]\) (c) \(e^{x}\left[(x-1) /\left(x^{2}+1\right)\right]\) (b) \(e^{x}\left[(2 x-2) /\left(x^{2}+1\right)\right]\)

If \(\int\left[\mathrm{dx} /\left(\sin ^{6} \mathrm{x}+\cos ^{6} \mathrm{x}\right)\right]=\mathrm{K} \tan ^{-1}[(\tan 2 \mathrm{x}) / 2]+\mathrm{c}\) then \(\mathrm{K}=\) (a) \((\overline{1 / 2)}\) (b) \(-1\) (c) 1 (d) \(-(1 / 2)\)

If \(\int \sin ^{3} x d x=A \cos ^{3} x+B \cos x+c\) then \(A-B=\) (a) \((4 / 3)\) (b) \(-(4 / 3)\) (c) \((1 / 3)\) (d) \(\overline{-(1 / 3)}\)
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