Right Triangle Method to solve integrals

[kidmo87]

Hello everyone. When my teacher was solving problems for integrals, she drew right triangles for each problem. Its been soo long that I cant remember. Can anyone point me to the right direction, or tell me a youtube video so I can remember how to do it. Here are some samples that she did to clarify what im talking about. Thanks.

 

[kidmo87]

sorry, maybe im not explaining this right. Like for the first right triangle, the purpose was to find out that x= 4 sin theta. The second right triangle to figure out that x=5sec theta. and the third right triangle was to figure out that x= tan theta. My question is by looking at the equations, how to solve in order to know what trig function it is? thanks.

 

[srmichael]

sorry, maybe im not explaining this right. Like for the first right triangle, the purpose was to find out that x= 4 sin theta. The second right triangle to figure out that x=5sec theta. and the third right triangle was to figure out that x= tan theta. My question is by looking at the equations, how to solve in order to know what trig function it is? thanks.

There are three main integrals that you need to know and then what your teacher did will make sense.

\(\displaystyle \int\frac{du}{\sqrt{a^2-u^2}}=sin^{-1}\frac{u}{a}+c\)

\(\displaystyle \int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}sec^{-1}\frac{u}{a}+c\)

\(\displaystyle \int\frac{du}{a^2+u^2}=\frac{1}{a}tan^{-1}\frac{u}{a}+c\)

 

[DrPhil]

Hello everyone. When my teacher was solving problems for integrals, she drew right triangles for each problem. Its been soo long that I cant remember. Can anyone point me to the right direction, or tell me a youtube video so I can remember how to do it. Here are some samples that she did to clarify what im talking about. Thanks.View attachment 2721

Is it possible your pictures are wrong? They don't make sense to me. What you show on the horizontal leg for the sine and secant should be the hypotenuse, and for the tangent the horizontal leg is a=1. Then the radical in each integrand presents a form of the Pythagorean theorem.

 

[kidmo87]

oh yea now I remember. Thanks a lot. I just found those formulas too. thanks.