Integration q (1 Viewer)

Life'sHard · Nov 21, 2021

Tan sub --> Change to sec in denominator --> Change dx into dtheta the rest should look pretty standard.

5uckerberg · Nov 21, 2021

=)(= said:
View attachment 34060

Life'sHard said:
Tan sub --> Change to sec in denominator --> Change dx into dtheta the rest should look pretty standard.

Do that and play with the problem.

=)(= · Nov 23, 2021

i get sin(theta) but the answer is

CM_Tutor · Nov 23, 2021

It is $\sin\theta$ , which you need to express in terms of $x$ using the fact that $x = \tan\theta$ .

icycledough · Nov 23, 2021

=)(= said:
i get sin(theta) but the answer is View attachment 34124

As said above, make sure to convert back to question form. Students make a common mistake whereby they use a substitution (commonly a 'u' sub or a 'trig' sub), but forget to convert it into x or whatever variable the question has used.

=)(= · Nov 23, 2021

i am not sure how to convert this back into that form

CM_Tutor · Nov 23, 2021

=)(= said:
i am not sure how to convert this back into that form

Approach 1:

$\text{Since $\tan\theta = x$ and taking $x > 0$, $\theta$ is an angle in a right-angled triangle where the two non-hypotenuse sides are $x$ and 1.}$

$\text{So, the length of the hypotenuse is $\sqrt{x^2+1}$ (applying Pythagoras' Theorem), meaning that $\sin\theta = \cfrac{x}{\sqrt{x^2+1}}$.}$

Approach 2:

$\text{Using the Pythagorean identity $\tan^2\theta + 1 = \sec^2\theta$, it is clear that $\sec^2\theta = x^2 + 1\ \implies\ \cos^2\theta = \cfrac{1}{x^2+1}$.}$

$\text{Now, rearranging and applying the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ produces}$

$\sin^2\theta = 1 - \cos^2\theta = 1 - \cfrac{1}{x^2 + 1} = \cfrac{1 + x^2 - 1}{x^2+1} = \cfrac{x^2}{x^2+1}$

$\implies\ \sin\theta = \cfrac{x}{\sqrt{x^2+1}} \quad \text{noting that the sign of $\sin\theta$ matches the sign of $x$.}$

Approach 3:

$\text{You have the answer down to $\sin\theta$ where $x = \tan\theta$, and so $\sin\theta = \sin{\tan^{-1}{x}}$, which can be simplified to $\cfrac{x}{\sqrt{x^2+1}}$.}$

Integration q (1 Viewer)

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