## 17Calculus Integrals - Inverse Trig

##### 17Calculus

When you can recognize terms in the integrand that are direct derivatives of inverse trig functions, you can immediately write down the integral. The key is to know the derivatives.

From derivatives, you can get these integral forms.

 $$\displaystyle{ \int{ \frac{1}{\sqrt{1-t^2}} ~dt } = \arcsin(t) }$$ $$\displaystyle{ \int{ \frac{1}{1+t^2} ~dt } = \arctan(t) }$$ $$\displaystyle{ \int{ \frac{1}{\abs{t}\sqrt{t^2 -1}} ~dt } = \arcsec(t) }$$

Although there are actually six inverse trig functions, remember that the remaining ones differ from these by just a negative sign.

When you get to calculus 2, you will not need these integrals in exactly these forms. You will learn a technique called trig substitution that will allow you to integrate not only these integrals directly but also more complicated integrals that are similar in form to these. Okay, try these practice problems.

Practice

Evaluate these integrals giving your answers in exact, simplified and factored form.

$$\displaystyle{ \int_0^1{ \frac{2x+1}{x^2+1} ~ dx } }$$

Problem Statement

$$\displaystyle{ \int_0^1{ \frac{2x+1}{x^2+1} ~ dx } }$$

Solution

### Michael Penn - 4134 video solution

video by Michael Penn

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