Let's do some calculus! (1)

Calculus Level 5

$\large \int{\dfrac{\csc^{2}x}{\sqrt{\cot^{2}x-1}}}\,dx - 2\int{\dfrac{\sin x}{\sqrt{1-2\sin^{2}x}}}\,dx$

Which of the given integrating functions is equivalent to the integration above:

$\begin{array} {} & \displaystyle \int{\sqrt{\csc^{2}x-2}}\,dx & \implies (2) \\ & \displaystyle \int{\dfrac{\sqrt{\cos 2x}}{2\sin x}}\,dx & \implies (3) \\ & \displaystyle \int{\dfrac{\sqrt{\cos 2x}}{\sin x}}\,dx & \implies (5) \\ & \displaystyle \int{\sqrt{\csc^{2}x-1}}\,dx & \implies (7) \end{array} $

Details:

Multiple options may be correct.
Find out the answer as the product of the numbers corresponding to the correct integrating function(s), then type in your answer as the logarithm of the product at base $10$ .
The corresponding numbers are the prime numbers in parentheses at the end of each option.
Example: If there were three correct answers with corresponding numbers $5,17,23$ , then the answer would be $\log_{10}{\left(5\times17\times23\right)}$ or $\log_{10}{5}+\log_{10}{17}+\log_{10}{23}$ (whichever way necessary) which would be approximately equal to $3.291$ .
Give your answer correct to 3 decimal places.

For more problems on calculus, click here .

The answer is 1.

1 solution

Chew-Seong Cheong
Nov 1, 2017

$\begin{aligned} I & = \int \frac {\csc^2 x}{\sqrt{\cot^2 x - 1}} dx - 2 \int \frac {\sin x}{\sqrt{1-2 \sin^2 x}} dx \\ & = \int \left(\frac {\csc^2 x}{\sqrt{\cot^2 x - 1}} dx - \frac {2\sin x}{\sqrt{1-2 \sin^2 x}} \right) dx \\ & = \int \left(\frac {\sin^2 x \cdot \csc^2 x}{\sin^2 x \sqrt{\cot^2 x - 1}} dx - \frac {2\sin x}{\sqrt{1-2 \sin^2 x}} \right) dx \\ & = \int \left(\frac 1{\sin x \sqrt{\cos^2 x - \sin^2 x}} dx - \frac {2\sin x}{\sqrt{1-2 \sin^2 x}} \right) dx \\ & = \int \frac {1 - 2\sin^2 x}{\sin x \sqrt{1 - 2\sin^2 x}} dx \end{aligned}$

$\begin{aligned} \ \ & = \int \frac {\sqrt{1 - 2\sin^2 x}}{\sin x} dx & \small \color{#3D99F6} \implies A: 2 \\ & = \int \frac {\sqrt{\cos 2 x}}{\sin x} dx & \small \color{#3D99F6} \implies C: 5 \end{aligned}$

$\implies$ Answer: $\log_{10} (2\cdot 5) = \boxed{1}$

Isn't it true only when sinx is positive ? @Chew-Seong Cheong @Tapas Mazumdar

Ankit Kumar Jain - 3 years, 1 month ago

I don't think so.

Chew-Seong Cheong - 3 years, 1 month ago

While going to the fourth line form the third one in the solution , sir , you have sent sinx inside the square root , won't it be true only if it is positive?

Ankit Kumar Jain - 3 years, 1 month ago

@Ankit Kumar Jain – You can plot the integrands to find out. It is not real for all value of $x$ , because of $\sqrt{\cot^2 x - 1}$ and $\sqrt{1-2\sin^2 x}$ . But it should takes some $x < 0$ and hence $\sin x < 0$ .

Chew-Seong Cheong - 3 years, 1 month ago

@Chew-Seong Cheong – But then if sinx is negative then shouldn't we apply a minus sign before sending it inside the root .. I mean to say x = $\sqrt{x^2}$ only when x is positive.. Please clarify the flaw in my reasoning .( I am a beginner in calculus )

Ankit Kumar Jain - 3 years, 1 month ago

@Ankit Kumar Jain – I don't actually know what you are saying. But $0 \le \sin^2 x \le 1$ for all $x$ and therefore $\sqrt{1-2\sin^2 x}$ is defined for $-\frac \pi 2 \le x \le \frac \pi 2$ in the first quadrant, where $-\frac 12 \le \sin x \le \frac 12$ .

Chew-Seong Cheong - 3 years, 1 month ago

Let's do some calculus! (1)

For more problems on calculus, click here .

The answer is 1.

1 solution

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