Simplifying Trigonometric Antiderivative using Right Triangles?

[OmniNewton]

Mod note: Moved from a homework section
1. Homework Statement
Hello my question more has to do with theory that perhaps deals with algebra.

Why is the following true?

Homework Equations

N/A

The Attempt at a Solution

N/A[/B]

 

[fresh_42]

Have you tried to apply the formulas $cos^2x + sin^2x = 1$ and $tan (x)=\frac{sin (x)}{cos (x)}$?

 

[OmniNewton]

Have you tried to apply the formulas $cos^2x + sin^2x = 1$ and $tan (x)=\frac{sin (x)}{cos (x)}$?

Hello, fresh_42

Thank you for the reply I have attempted to apply the identity cos^2x + sin^2x = 1 and could not figure out an algebraic means to have this work out. As for the reciprocal identity tanx = sinx/cosx. I do not see how this identity applies to arctanx since arctanx does not equal arcsinx/arccosx

Regards,

OmniNewton

 

[Mark44]

Because you have integrals on both sides, you can instead show that the two integrands are equal (plus possibly a constant). IOW, just show that $\sin^2(\arctan(x)) = \frac{x^2}{x^2 + 1} + C$

The best way to do this, IMO, is to draw a right triangle, and label an acute angle $\theta$, with the opposite as x and the adjacent side as 1. The $\tan(\theta) = \frac x 1$. Now find $sin(\theta)$.