Evaluate the integral.
-Use the formula $\int f ^ { - 1 } ( x ) d x = x f ^ { - 1 } ( x ) - \int f ( y ) d y , y = f ^ { - 1 } ( x )$ to evaluate the integral. $\int \cos ^ { - 1 } x d x$
A) $x \cos ^ { - 1 } x - \sin \left( \cos ^ { - 1 } x \right) + C$
B) $x \cos ^ { - 1 } x - \sin x + C$
C) $x \cos ^ { - 1 } x + \sin \left( \cos ^ { - 1 } x \right) + C$
D) $x \cos ^ { - 1 } x - x + C$

Question

Quizplus · Accepted Answer

We use the formula $\int f ^ { - 1 } ( x ) d x = x f ^ { - 1 } ( x ) - \int f ( y ) d y , y = f ^ { - 1 } ( x )$, with $f(x) = \cos(x)$ and $y = \cos^{-1} x$, or equivalently, $x = \cos y$. Then, $dy/dx = -1/\sqrt{1-x^2}$, so $dx = -\sqrt{1-x^2}dy$. Substituting into the formula, we get:
$$\int \cos ^ { - 1 } x d x = x \cos ^ { - 1 } x + \int \sin y (-\sqrt{1-x^2}) d y$$
Now, we use the identity $\sin(\cos^{-1}x)=\sqrt{1-x^2}$ to simplify:
$$\int \cos ^ { - 1 } x d x = x \cos ^ { - 1 } x - \int \sin(\cos^{-1}x) d y$$
Substituting back $x = \cos y$, we get:
$$\int \cos ^ { - 1 } x d x = x \cos ^ { - 1 } x - \sin(\cos^{-1}x) + C$$
Using trigonometric identities, we know that $\sin(\cos^{-1}x) = \sqrt{1-x^2}$, so the final answer is:
$$\int \cos ^ { - 1 } x d x = x \cos ^ { - 1 } x - \sin(\cos^{-1}x) + C = x \cos ^ { - 1 } x - \sqrt{1-x^2} + C$$
Therefore, the best choice is (A).

Evaluate the Integral ∫f−1(x)dx=xf−1(x)−∫f(y)dy,y=f−1(x)\int f ^ { - 1 } ( x ) d x = x f ^ { - 1 } ( x ) - \int f ( y ) d y , y = f ^ { - 1 } ( x )∫f−1(x)dx=xf−1(x)−∫f(y)dy,y=f−1(x)

Evaluate the Integral $\int f ^ { - 1 } ( x ) d x = x f ^ { - 1 } ( x ) - \int f ( y ) d y , y = f ^ { - 1 } ( x )$