Question

solve the question fast​

Answer 1

Answer:

its answer is 49

please give me points

Answer 2

Answer:

7. $2\theta\sin\theta+ \theta^2\sin\theta - \csc^2\theta$

8. $\int_{1}^{2}\left(\frac{1}{x}- e^{-x}\right)$ = $ln2 + \frac{1}{e^2} - \frac{1}{e}$

$\int^2_{1}\frac{\ln x}{x}\, dx=\frac{3}{2}$

Step-by-step explanation:

The explanation for Answer 7 is as follows:

$r = \sec\theta\csc\theta - \tan\theta+\theta^2\sin\theta\\\frac{dr}{d\theta} = (\sec\theta\tan\theta)(\csc\theta) + (\sec\theta)(-\cot\theta\csc\theta) - \sec^2\theta + 2\theta\sin\theta + \theta^2\cos\theta\\\frac{dr}{d\theta} = \frac{1}{\cos\theta}\times\frac{\sin\theta}{\cos\theta}\times \frac{1}{\sin\theta} - \frac{1}{\cos\theta} \times \frac{\cos\theta}{\sin\theta} \times\frac{1}{\sin\theta} - \frac{1}{\cos^2\theta} + 2\theta\sin\theta+ \theta^2\cos\theta$

$\frac{dr}{d\theta} = \sec^2\theta - \csc^2\theta - \sec^2\theta + 2\theta\sin\theta+\theta^2\sin\theta\\\Rightarrow 2\theta\sin\theta+ \theta^2\sin\theta - \csc^2\theta$

Formulas use:
Derivative of

$\sin\theta = \cos\theta\\\cos\theta= -\sin\theta\\\tan\theta = \sec^2\theta\\\sec\theta = \sec\theta\tan\theta\\\csc\theta = -\csc\theta\cot\theta\\\cot\theta = -\csc^2\theta$

The explanation for answer 8 is as follows:
$\int^2_{1}\frac{\ln x}{x}\, dx\\\text{Let lnx = t}\\\frac{dt}{dx} = \frac{1}{x}\\dt = \frac{dx}{x}\\\int_{1}^{2} t\,dt = \frac{t^2}{2}]^{2}_{1} = \frac{4-1}{2} = \frac{3}{2}$

$\int_{1}^{2}\left(\frac{1}{x}- e^{-x}\right)\, dx\\\int_{1}^{2} \frac{1}{x}\, dx - \int_{1}^{2}e^{-x}\, dx\\\ln x]^{2}_{1} - (-e^{-x})]^{2}_{1}\\\ln2- \ln1 + e^{-2} - e^{-1}\\\ln2 + \frac{1}{e^2} - \frac{1}{e}$

I hope my answer is helpful.